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Text
***** V KITOB *****
#1#
01/11/2015 у
***** V KITOB *****
РА1Ш112Т1Ш CIIIZIQU TPiXGLAMALAll.
1. Parametrli chiziqli tenglamalarni yeching.
1) m ning qanday qiymatlarida m(mx-1) = 9* + 3 tenglama cheksiz
ko’p ildizga ega?
2) m ning qanday qiymatlarida m(mx-1) = 4* + 2 tenglama cheksiz
ko’p ildizga ega?
3) m ning qanday qiymatlarida m(mx-l) = 16x + 4 tenglaffi||hsksiz||
ko’p ildizga ega?
V VKVWW’ANV
4) m ning qanday qiymatlarida m(mx - \) = 25x + 5 tenglama cheksiz
ko’p ildizga ega? J|
5) m ning qanday qiymatlarida m{mx-\) = 36x+6 tengl||na cheksiz
ko’p ildizga ega?
6) m ning qanday qiymatlarida m(mx-l) = 49*- 7 tenglama cheksiz
ko’p ildizga ega?
^ ” '‘^/XvXvivIvIvIviviv^
7) m ning qanday qiymatlarida m(mx-1) Ц 81*-9 tenglama cheksiz
■mSk ■
-1) = 100л: -10 tenglama cheksiz
•ХХХуУ.уХу
ko’p ildizga ega?
..;>>< \s\<wy:y:^
8) m ning qanday qiymatlarida Щ
•mx *•. Л "t1*
ko’p ildizga ega?
г о о W
9) m ning qandav qivmatlaadlb/wy+l = m tenglama yechimga ega
, ,, ,. lillx
bo lmaydi.
V.SS'.'H.'.V.S’.SSS
S'tS'.V.V.V.SNS*.*.
•.'.•.•.V.V.V.'.'.V.S
V.’.'.VAV.V.-.VA
X» .«. - - • ■ . •
'.VAWASVAV. V.V
10) m ning qanday qiymatlarida my-5 = m tenglama yechimga ega
bo’ lmaydi,.... Vr
11) m n(ng qanday qiymatlarida (m - 2)y + 4 = 3m tenglama yechimga
•XvX*x*. vXv.'Xv
{m + 6)y — 13 = 8m tenglama yechimga
* vvSSJbvv‘,v ‘ 2
13) m ning qanday qiymatlarida (m - 4)y -8 = 5m tenglama yechimga
ega bo’lmaydi.
2
14) m ning qanday qiymatlarida (m -9)y-2\ = 7m tenglama
yechimga ega bo’lmaydi.
2
15) m ning qanday qiymatlarida (nr - 36)y-54 = 9m tenglama
yechimga ega bo’lmaydi.
16) a ning qanday qiymatlarida ax-a -x-1 tenglama cheksiz ko’p
yechimga ega bo’ladi?
SOLIYEV X. X.
#2#
03/11/2015 у
***** у KITOB *****
17) п ning qanday qiymatlarida nx+\ = n + x tenglama cheksiz ko’p
yechimga ega bo’ladi?
18) n ning qanday qiymatlarida nx + 3 = n + 3x tenglama cheksiz ko’p
yechimga ega bo’ladi?
19) n ning qanday qiymatlarida nx-2-n-2x tenglama cheksiz ko’p
yechimga ega bo’ladi?
20) n ning qanday qiymatlarida nx-1 = n-1 tenglama cheksiz ko’p
yechimga ega bo’ladi?
21) m ning qanday qiymatlarida m2x
yechimga ega bo’ladi?
22) m ning qanday qiymatlarida m2x -m = 25x + 5 tenglama cfiiftz
ko’p yechimga ega bo’ladi?
23) m ning qanday qiymatlarida m x-m- 3(&x,-46 tenglama cheksiz
-m =
M
V.W.'
n’.VAV
л +1 tenglamsfbpksiz ko’p
v! vl'i v\‘!v!v. v. v.%
• ^A\VA44y>,SN'\V.V.'
• *^.V.V.-ASVWA'.V.SV
1 ЛУЛ'.*.'. й
AWA'A.
ko’p yechimga ega bo’ladi?
24) m ning qanday qiymatlarida m x-gfci = 81л-9; tenglama cheksiz
ko’p yechimga ega bo’ladi?
Як. \
JjXwvavaja, , >X;
•X\W
25) m ning qanday qiymatlarid^^Mfc;|F 16л +12 tenglama cheksiz
ko’p yechimga ega bo’ladi? Jjk %
26) m ning qanday qiyma|jari^|||“f ;:“iil,w = 64л-56 tenglama cheksiz
« * « • « * 4 . « • Л V.-.
/л5пх/^1тгто orro
ko’p yechimga ega bo’ladi? 4
//!v!'!v!v!s‘v!vava .
VA V.y .|A\\\N V. VAV.V..^
27) m ning qanday qiymatlarida m x-5m = 4л+ 10 tenglama cheksiz
VA'A’.V.'.V.'AV
4\\\v.v.v
ko’p yechimga egaBiladi?
%%улулу»улу» v»^
28) a ning qanday qiyrlltlarida ax = 2x + 3 tenglama cheksiz ko’p
yechimga egifll|ladi?v^r
\j,vav.v!va|Xs^ \\\*^ v!y
29) a ning!|anday :qiyijiatlarida ох = 5л + 11 tenglama yechimga ega
bo’lmaydi? !1|||
30)||| nirig qanday qiymatlarida ax = -Зл + 7 tenglama yechimga ega
!\v!v!'t.4v!v.v!*
31) a ning qanday qiymatlarida ax = 4л -9 tenglama yechimga ega
bo’lmaydi?
32) a ning qanday qiymatlarida ax - 3 = a + 2л tenglama yechimga ega
bo’lmaydi?
33) a ning qanday qiymatlarida ax+ 7 = a-5л tenglama yechimga ega
bo’lmaydi?
34) a ning qanday qiymatlarida ax-9 = 3a + 7x tenglama yechimga ega
bo’lmaydi?
35) a ning qanday qiymatlarida Зал + 8 = 2а-15л tenglama yechimga
ega bo’lmaydi?
SOLIYEVX.X.
#3#
01/11/2015 у
***** у KITOB *****
36) a ning qanday qiymatlarida 7ax-\3 = 5a + \4x tenglama yechimga
ega bo’lmaydi?
37) n ning qanday qiymatlarida nx + 5 = n-2x tenglama yechimga ega
bo’lmaydi?
38) n ning qanday qiymatlarida л*-8 = 5л + 3* tenglama yechimga ega
bo’lmaydi?
x*x*>
л
39) п ning qanday qiymatlarida n x - 7 = 9n - Ax tenglama yechimga
ega bo’lmaydi?
2 ...
40) n ning qanday qiymatlarida n~x-3 = n-9x tenglamall^echimga ega
bo’lmaydi?
41) Ushbu (a2 - \)x + 3 = 0 tenglama yechimga ega bo’lmaydiganli
ning barcha qiymatlari yig’indisini hisoblang.
... ” 4ft
42) Ushbu {a -9)*-5 = 0 tenglama yechimga ega bo’lif|ydigan a
ning barcha qiymatlari yig’indisini hisoblang.""1""
, >V* ‘•■•Xvlvlvlvlv/Iv/
43) Ushbu (a2 - 25)x + 8 = 0 tenglama yechimga e|a Jb’lmaydigan a
ning barcha qiymatlari yig’indisini hisoblang.
44) Ushbu {a2 - 49)*-14 = 0 t^famMlIhirnga ega bo’lmaydigan a
ning barcha qiymatlari yig’i
45) Ushbu {a2 - 64)* + 15 = 0 tenglania yechimga ega bo’lmaydigan a
ning barcha qiymatlari yig’indisini hisoblang.
46) n ning qanday qiymatlaridi|f? (y-l) = y-n tenglamaning ildizi
.V.W-
47)
yo’q? .
n ning q^J|y qiynfffllrida n~{y-\) = 9y-3n tenglamaning ildizi
yo’q?
C-y.v.y
n
A*.
AVA'.VW.V.VA AV,
-iv:-
ЩШ+ШХ:,
48) n ning qanday qiymatlarida n {у -1) = 25^ - 5n tenglamaning ildizi
..v;*:^Xs\4
49) ;Щ|щ qanday qiymatlarida n (y-\) = 36y + ln tenglamaning
\V.V>.4V.4V.V
50) n ning qanday qiymatlarida n 0м-1) = 49у-8и tenglamaning ildizi
yo’q?
51) n ning qanday qiymatlarida n (y + 3) = 81y + l \n tenglamaning
ildizi yo’q?
52) Tenglama a ning qanday qiymatlarida cheksiz ko’p yechimga ega?
10(ax -X) = 2a-5x-9
SOLIYEVX.X.
#4#
01/11/2015 у
***** V KITOB *****
2
53) a ning qanday qiymatida (a + 2)x = a(x -a) + 2 tenglamaning
ildizlari cheksiz ko’p bo’ladi?
54) a ning qanday qiymatida (cr - 2)x = a(x-a) + 4 tenglamaning
ildizlari cheksiz ko’p bo’ladi?
2
55) a ning qanday qiymatida (a - 6)x = a(x -a) + 9 tenglamaning
ildizlari cheksiz ko’p bo’ladi?
56) a ning qanday qiymatida (cr -56)* = a{x-a)+ 64 tenglamaning
ildizlari cheksiz ko’p bo’ladi?
57) a ning qanday qiymatida (a2 -12)* = a(x-a)-25 j
y.'/.'.’.V.W.y.V.V.
•w.w •
4\V.S\*AV.V.V.y.V.*.V.,A
■ • мг' wa\n\v.\: 111 • e g
ildizlari cheksiz ko’p bo’ladi?
58) a ning qanday qiymatida (cr - 20)* = a(x - a) + 36 tenglaiianing
ildizlari cheksiz ko’p bo’ladi?
59) Tenglama a ning qanday qiymatida yechimg§; eg^ emil?
1) 6x-a-6 = (a + 2)(* + 2) ; 2) 3*-8 = (a-5|fp-1) ;
3) 2x-4a + 9 = (2a- 7)(*-3) ; 4) ;5*:- 2q +11 = (3a +1)(* + 2) .
60) Tenglama к ning qanday qiy|pa!i!i|lbchimga ega emas?
2kx + 3 к-2 + х 2k-5 + 2x
D-
3)
3
A* + 8
лу.улчч;
3k + 5 + 7*
5-
Ь 'v ж. A
L-T
ШШЛк + 1-Зх
KwiwXy'
A*.V.\WAV.W.
V.y. Ч'ЛЧ'.'.'.Л'ЛЧ
6 8
61) к ning qanday 5l||||||tida
12 15 • i
+ 6)x = k + 7(* +1) tenglama yechimga
ega emas?
. ;vlv!v
чЁШР
62) к ning qanday qiym|lida k(k -4)x = k + 5(* -1) tenglama yechimga
63) Jk ning qanday qiymatida k(k-7)x = 3k-\2(* +1) tenglama
ning qanday qiymatida k(k + 3)x -2k + 4(*-1) tenglama
|ga emas?
65) a ning qanday qiymatida
3* - a ax-4
tenglama yechimga ega
emas?
2* + 5a _ ax - 3
66) a ning qanday qiymatida ~ - ^ tenglama yechimga ega
emas?
SOLIYEV X. X.
#5#
01/11/2015 у
***** V KITOB *****
67) a ning {cr -4)* + 5 = 0 tenglama yechimga ega bo’lmaydigan
barcha qiymatlari ko’paytmasini toping.
68) a ning (a -9)*-15 = 0 tenglama yechimga ega bo’lmaydigan
barcha qiymatlari ko’paytmasini toping.
2
69) a ning {a -36)* + 7 = 0 tenglama yechimga ega bo’lmaydigan
barcha qiymatlari ko’paytmasini toping.
2
70) a ning (<r - 64)* - 24 = 0 tenglama yechimga ega bo’lmaydigan
barcha qiymatlari ko’paytmasini toping. ж%
71) a ning (a2 - 81)* +13 = 0 tenglama yechimga ega boslm|ydigan
.\W.\4\4
'll
AWAWA .
A*.V.V.,.,A*.,.,A ,
A'A’AVAWAVXV.
AWAWA’AWAWAV
ytfAWA’AWA'A’A’AWA .
AW.VAVAVA'AWA’AV.W.s
VAWAAWAWA'.VAWA'AVA .
... VVVAVAV.'.V/.VAVV.VAWi.
л‘.\чЛ .
barcha qiymatlari ko’paytmasini toping.
2
72) a ning О" -100)* -25 = 0 tenglama yechimga ega bo’lmaydigan
barcha qiymatlari ko’paytmasini toping.
73) Ushbu nx = nA -12 tenglamaning ildizlarflif|i|||qp Bo’ladigan
n(n e N) ning barcha qiymatlari yig’indisini toping.
74) Ushbu nx = n -36 tenglamaning ildizlari natural son bo’ladigan
«(« e N) ning barcha qiymatlariipig’mllllitoping.
75) Ushbu nx = n2 -48 tenglainaning ildizlari natural son bo’ladigan
n{n e N) ning barcha qiymatlari yig’ indisini toping.
x;:- ^
^ • J.vXwXv,
76) Ushbu wc = n~ - 56 tenglamaning ildizlari natural son bo’ladigan
n(n e N) ning barcha qiymatlari yig’indisini toping.
2 SV.j.VAV.V.V.j.^ '.y.y.V.y
77) Ushbu nx^n -64;tenglamaning ildizlari natural son bo’ladigan
;;л*й'Й*л t
n{n e N) ning barcha qiymatlari yig’indisini toping.
78) Ushblfjyg^SCTtenglamaning ildizlari natural son bo’ladigan
Xv!;Xv!vX;!;!va XvIvIvXvav!;.
«(« € iV) ning barcha qiymatlari yig’indisini toping.
•>X *w*X*X'X\\\4v. ^ ‘vwssMs,!< • • Ж •/ •' ^ ^
V.S SS V.VA,A,A,.<
лv.'Ww'W \\ v.sv.v.w
;щ=/г -96 tenglamaning ildizlari natural son bo’ladigan
m
ъхШ&ъШ.....
■ vav«KSswa v. •. v •
ning barcha qiymatlari yig’indisini toping.
_V|V. ^. > • • • • VAV.V, 2
80) TJshbif wx = n -100 tenglamaning ildizlari natural son bo’ladigan
n{n e N) ning barcha qiymatlari yig’indisini toping.
81) Ushbu 10(ax-l) = 2a-5x-9 tenglama a ning qanday qiymat-larida
yagona yechimga ega?
82) Ushbu 3(ax + 4) = 5a - 6x + 2 tenglama a ning qanday qiymat-larida
yagona yechimga ega?
83) Ushbu 6(ax-3) = 7я + 12х-1 tenglama a ning qanday qiymat-larida
yagona yechimga ega?
SOLIYEVX-X-
#6#
01/11/2015 у
***** у KITOB *****
84) Ushbu 2(ax+5) = a + Sx + 7 tenglama a ning qanday qiymat-larida
yagona yechimga ega?
85) Ushbu 7(ax-2) = 9a-x + \\ tenglama a ning qanday qiymat-larida
yagona yechimga ega?
6x-m Imx +1
86) m ning qanday qiymatida
nolga teng bo’ladi?
87) m ning qanday qiymatida
nolga teng bo’ladi?
88) m ning qanday qiymatida
nolga teng bo’ladi?
89) m ning qanday qiymatida
nolga teng bo’ladi?
90) m ning qanday qiymatida
2
3 L'
2x + 5m
лгис — 12
6
8
5x - 3m
2mx-l
5
6 ftl|
•AWA V:
4 x + m :
wXwXvXvXv
,.'|',V.VAV.,.V.VAV.*.V
1тхТШ
12 .Щ
15 1
I
sNVA'.VA
V'A'A'.VAW.
s*A’AAWA-a\.
V.’.’.'.'.V.V.V.V.W.
VA’A’.VA'A'AV.V.VA.
4WAVAV.V,V,\%V,V.
%
VA’A.VA’AW.VAAWA.
•Л'.уЛ’.у.уЛ'Л’Л'ЛЧЧ. 4
_ ч'.улчул'лчу.ул'л*. —
./.•.v. ^ • •AW.VA'AWAW.yA Л
лш5тщ Шс 3
V • _ .
tenglamaning ildizi
тшжтжь.
’•'•У.
II
Щбл.,.
. • •v.sy.
Ay. •
ij.y.y.y
HI
ildizi
.v.;
Щг tenglamaning ildizi nolga
W-Й*, A*.
teng bo’ladi?
91) a ning qanday qiyniltianda = 6*-4 tenglama bitta musbat
yechimga ega?
92) a ning qanday qiymatlarid§li*(2* - a) = Sx -16 tenglama bitta musbat
ft ‘
* \v!*.v!v!vX'!v!v!s
'.S'.S'.S'A'.vXvV.S
SWA'A'iSSyA'A . .*.S
S'ASVAWAWAW.
•iV.SiSWAW.V.*
Ж’.'.-в'.'Л'А'Л'/Л a
|bJ
4N* 'AVAVA*AWA\
AV.'AV.V
93) a ning qanday qiymatlarida a{x - a) = 3x-9 tenglama bitta musbat
94) a
v.-.v.v.v.v.v.s;.
A’A'AVAVA
ega emas?
2) 2ax-3 = &x + 6b ; 3) 5ax-12 = 13.x-26 ;
\b ; 5) 3ax + 18 = 1 5jc — 3Z> ; 6) 9ax-6 = 21 x + 4b.
/asnvW'.n’.vansvXva'.va-A'a 7 / 7 /
S\\SVA*A4SNSWA4SW.V4\,.VAV
^NSSWAVNNVA'A'A'AWAWAN
95) Ushbd;(£' -4k + 2)x = k-x-3 yoki (k + 2)x-\ = k + x
V.y.y.V.s .ч’Л
tenglamalardan biri cheksiz ko’p yechimga ega bo’ladigan к ning nechta
qiymati mavjud?
96) к ning (к2 - Зк + \)x = к - x - 4 va (к + \)x +1 = к + x tenglamalardan
hech bo’lmaganda birining cheksiz ko’p yechimga ega bo’ladigan nechta
qiymati mavjud?
97) к ning qanday qiymatlarida /;(* + l) = 5 tenglamaning ildizi musbat
bo’ladi?
SOLIYE V X- X.
#7#
03/1V2015 у
***** V KITOB *****
98) a ning qanday qiymatlarida ax -2a = 2 tenglama birdan kichik
ildizga ega bo’ladi?
99) b ning qanday qiymatlarida b(2 -x) = 6 tenglamaning ildizi manfiy
bo’ladi?
100) k ning qanday qiymatlarida tenglama manfiy yechimga ega
Ax -1
bo’ladi?
= k + 3
x -1
101) m ning qanday qiymatlarida
4-m =
x -1
»SVS*AV
tenglamp|n| ildizlari
. »^.4\\\nvv!v.v.v.4\ -■
w.y
musbat bo’ladi?
102) t ning qanday qiymatlarida 3x - A = 2(x -1) tenglama mulbaf
ildizga ega?
103) a ning qanday qiymatlarida 3{x +1) = 4 +a* tenglamaning ildizi
-1 dan katta bo ladi?
104) к ning qanday qiymatlarida
ega?
= к-2
:;Г
tenglama manfiy ildizga
.S‘.VA*.
v.ww.v.vwvv
-ч * \v ’
105) Tenglamaning ildizlari manfiy, bo’ladigan к ning barcha butun
•AWAW.V.N'.-. .
», WAW.V.V.SV, Ч'Л'Л'.У
;.vv w.y.v.v.v.v.s
_ --.VVV* . •.v.v.v.sw.y.s
• /.‘ЛТ. • ГЛ'А'Л'Л'.'Л'.'Л
•А’.ч
~2)2 -у = к2 -25
musbat qiymatlari yig’indisini toping.
106) p ning qanday qiymatlarida 4л - 7p = 4 tenglama musbat ildizga
ega?
107) p ning qanday qiymatla:ipa 5x-3p = A tenglamaning ildizi-12
dan katta bo’
v
108) p ning
ega? JjL
svsvsvvvsvsvsvsvs L
VS" S'.'.'AVy.VASVA
ЩЯ VV VV.Vvvsv.vv.v
.V.V.V.VVV.
VAVV'Ky.'. ^
AV.V.V.VT.V.
ШШШ-’ <y<-
' qiymatlarida 4л- Ip = 5 tenglama manfiy ildizga
109) filing fanday qiymatlarida lx + \2p = 7 tenglama manfiy ildizga
.vvXvsvsvsv •.vvr.vv.vv.vvvv
VSSSVSSSVV.SV VVVVVVVVVVVVVVV
•VXvXvvvXvvvvv.
i7 %
qanday qiymatlarida 5x-3p = 6 tenglamaning ildizi 8 dan
,\\\V4vXvvvsv*Xs****^*‘-‘\vsvsv\
‘.V.V.V.4 . . .NV
VSVVVSVSVVVSV
111) pining qanday qiymatlarida 6x + Sp = 7 tenglama musbat ildizga
ega?
3b + 2_
112) у”’” “ 2b tenglama b ning qanday qiymatlarida manfiy
yechimga ega bo’ladi?
113) a ning qanday qiymatlarida 3(x +1) = 4 + ax tenglamaning ildizi
-2 dan katta bo’ladi?
SOLIYE V X. X.
#8#
01/11/2015 у
***** у KITOB *****
114) * ning qanday qiymatlarida
yechimga ega bo’ladi?
4x-l
x-\
--k + 2 tenglama manfiy
1. Parametrli tenglamalar sistemasini yeching.
WV\Wi*
• * * VAN
%
1) к ning qanday qiymatlarida sistemaning birorta ham yechimga egalll
bo’lmaydi?
** NN'.'.NN'.NNN'.V.V.V.V.VANVA*. .
* VANVANVANVANVANN AVAV. -
* ’ NV.'AVANVANNWAVAVA'A ч
■.•.•.'.'.•.•.NW.V.'AVAV.NV.NV.**
(к2-k-\)x + 2,5y-5 = 0
2x + y + к = 0
1)
2) a ning
bo’lmaydi?
ax- у = 0
x + у = 10 ’
f x + ay = l
[ax + у = 2a
2)
Uk2 + к + l)x + 3y - 6 = 0
x i.3% к = 0
&
■k . *
W"
m
W
• ■.V.N'AWA'
- v.4W —
'2k+ ay = 2 |||
5o№
»*•*%
|§| + 2y = 3
5)
Ua - 5 (ax-3y = 3
' 61 , 7гЦ- Hill fc 7|,i.fffl-7'
M; % ’ЩШч У
3) Tenglamalar sistemasilmAing qajftday qiymatlarida yechimga ega
bo’lmaydi?
mx + 2у + 4 = 0N|!||!|J 4xipiy + 9 = 0 \nix-у = 2
^ 12x + my - 8 = 0 ’ Щщ$х - 9у -11 = 0 ’ \x-my = 5 ’
к ningiq^d5|||ym|tlarida tenglamalar sistemasi cheksiz ko’p
як
ANVAVV.
•.NV.V.NV.
:a\nV
шхтш&и
Av w
ANNNV.S
V.NNNVA
.SV.VSVVN
......
NyXvXlXlXlXy'XlI’
\3x + (k -\)y = k + \
2)
3)
L (k + \)x + y = 3
ieSgljanday qiymatlarida tenglamalar sistemasi cheksiz ko’p
yechirqgalga bo’ladi?
(6 + a)x - 6 у = 2
- 2 ax + 3y = a-3
6) к ning qanday qiymatida tenglamalar sistemasi yagona yechimga ega
bo’ladi?
kx + 4y = 4
Зх + у = 1
kx-3y = l
5x + ky = 9
2x-y = 3 ’ \lx-2y = -3
—v4)
3x-6y = 4
x + ky = 2
#9#
SOUYEV X. X.
01/1V2015 у
***** V KITOB *****
7) a ning qanday qiymatida tenglamalar sistemasi yagona yechimga ega
bo’ladi?
1)
x + y = a Jx + y = a \x + y = a I x + y = a
xy = 9 ; 2')\xy = l6 ; 3){.xy = 25 ; 4){^ = 81 *
8) k parametming qanday qiymatlarida tenglamalar sistemasi yechimga
ega bo’lmaydi?
kx-3y = 6 \kx-5y = \0 \kx + 2y = % f x-ky = 2
О 1 Ov-i.-o i 2) i i 3) i 7v„_7 ; 4) 15x_2v = 9-
2x-y = 2 , [ 3x-y = 3 ’ 7 [ 7;c + _y = 7
9) a va Z> ning qanday qiymatida tenglamalar sistemasi filMiffiega
bo’lmaydi?
6л:-15>^ = b
2) '
К
1)
4)
ax-5y = -1
6* +15y = 6 +3
4x-6y = b
ax-9y = 4 ’
4*5 * -6y = b
5)
4*-ay = 12
Зл:-2Ду = Z>
- ay =15
тжтшж
:*v.
•W'
ax-5y = 4
10) a va b ning qanday qiymatida;||i|j|iialar sistlmasi cheksiz ko’p
yechimga bo’ladi?
Swww
x:*x%
1)
4)
3* - 4,5jy = b
2x-ay-4
3x-by = \ 0,5
ax-6y = 1
(ax-At
чЖ ~ Ж
.S'A’.S’.'/.V.'.'.N'.S * V\
. •■•AWA'AV/.VA'.W.V.S*,
+ 2,5^ = b
4x -ay = 8
• v.4VA4V»w.eAVA\4v»s
• VS'A'AW.V.VA'A’AW.
•.ЧЧЧ'ЛЧ'Л'Л'Л'.'Л
. .4%
■wAW.?X^
л;.у
, к ning qanday qiymatida x + у = 2 tenglik
.лчччулч
= 8 bo’Isa, k ning qanday qiymatida x +у = 2
■штщшщк , 51 ,.0
tenghk o rmli bo ladi?
О--:-:-.-- Ж
13) %£Ш| qanday qiymatida tenglamalar sistemasi yechimga ega?
x-ly-\
2x-4y-2
x + 3y = a
14) Tenglamalar sistemasi x = 3 va y = 2 yechimga ega bo’lsa, a va b
ning qiymatini toping.
1)
2x + 3y = 5
2*-5_y = 3
< x-y=2 .
2)'
x-2y = 2 . ^ <
x + 4y = a
lx-3y-a
К * V
SOLI YE V X. X.
#10#
Ol/U/2015 у
***** V KITOB *****
ax + by = 3
bx + ay = 2
15) Agar - * = 2 va я > 0 bo’Isa, Tenglamalar sistemasini yeching.
1)
4)
y2 - x2 = 8a
«
у + x = a2
y2-x2 =\Sa
y + x = a2
2)
5)
y2 -x2 = 10a
y + x-a2
y2-x2= 24a
у + x = a2
\y2-x2=16a
3) I y + x = a2
\y2 -x2 =32a
6)
y + x-a
M
%
Шш
iiiiiiiiii
ОШШ
РА1Ш1КТ1Ш KYADllAT T12NGLAMALA11.
r
1. Parametrli kvadrat tenglamalarni yeching.
•.ччччч*.ччч*.чч'.ччч\ .
• • • VW/AV.V.W.V/
II
1) Tenglamaning ildizi 0 ga teng bo’ladigan m
*4V***vIv V
ko’paytmasini toping, x2 -9x + (m2 - 4)(m2 - 9) =
. vi-.
2) Tenglamaning ildizi 0 ga teng bo’Tadjgan m ning barcha qiymatlari
ko’paytmasini toping. *2 +
:2
#-16)=0
3) Tenglamaning ildizi 0 ga:ten^|^’Ildigan m ning barcha qiymatlari
ж
-i) = o
ko’paytmasini toping. ЩжЖх +
•IvXvXvl v! v!;X v!vIv/X;!v.pt
4) Tenglamaning ildizi 0 ga teng bo’ladigan к ning barcha qiymatlari
ko’paytmasini topirig|>i2 + 3x + (k - 3){k2 - 36) = 0
.sS' ’vl’iv.'vl'.'v'.vl'Iv. w
Л'ЛУ V.-A'.SW.SWAV
Л'.ул ’.V.yAV.W.V.V. _
+ШШ.. О
■шштш 2 ** l
5) Ъ ning qanday qiymatiga ^ + ~х + b uchhad to’la kvadrat bo’ladi?
.ЧЧЧЧЧЧЧЧЧЧЧ
3
6) b ning qanday qiyffiatida x* +—x + b uchhad to’la kvadrat bo’ladi?
"l| J
b ning tjilday qiymatida * + bx+ 16 uchhad to’la kvadrat bo’ladi?
8) b ning qanday qiymatida *2 -12* + 6 uchhad to’la kvadrat bo’ladi?
9) к nixig qanday qiymatlarida *2 + 2(k - 9)x + k2 + 3k + 4 ifodani to’la
kvadrat shaklida tasvirlab bo’ladi?
A
10) Ushbu * - 6x + q = 0 tenglamaning ildizlaridan biri 2 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
11) Ushbu * -10* + q = 0 tenglamaning ildizlaridan biri 2 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
12) Ushbu * + 3* + q = 0 tenglamaning ildizlaridan biri 3 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
SOU YEV X. X. # 11 # 01/11/2015 у
***** V KITOB *****
13) Ushbu * + 9x - q = 0 tenglamaning ildizlaridan bin 5 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
14) Ushbu x2 - px + 8 = 0 tenglamaning ildizlaridan biri 4 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
15) Ushbu x2 — px + 15 = 0 tenglamaning ildizlaridan biri 3 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisini toping.
16) Ushbu x - px- 21 = 0 tenglamaning ildizlaridan biri 7 ga teng. By
tenglamaning barcha koeffisiyentlari yig’indisini toping^lllllib 1
17) Ushbu x - px- 24 = 0 tenglamaning ildizlaridan biri 8 ga teng. Bu
tenglamaning barcha koeffisiyentlari yig’indisinytoping.
V.V’iv
A .y.y.y.v
18) Ushbu x + px-12 = 0 tenglamaning ildizlaridan biry| ga teng. Bu
.ч\ч\чч<чч\чч<чулл>*.в ivIvlvXvl
tenglamaning barcha koeffisiyentlari vig’indiimlligin^. w
'■ \4y!v.v!y!v!v!v!v!v.v.v.y
19) Ushbu x2 + px — 42 = 0 tenglamanihg ildizlaridan biri 7 ga teng. Bu
tenglamaning barcha koeffisiyentlarillig’indisini toping.
20) p ning qanday qiymatida tenglamaning ildiz-laridan
biri 5 ga teng bo’ladi?
21) p ning qanday qiymaticfa *2 - px + 18 = 0 tenglamaning ildiz-laridan
biri 3 ga teng bo’ladi?
22) p ning qanday qiymatida x2 + />*-28 = 0 tenglamaning ildiz-laridan
biri 4 ga teng bo’ladi?-
Ь M ЩШЩШ:, W 2
23) p ning qanday qiyi||Ma x - px +12 = 0 tenglamaning ildiz-laridan
biri 2 ga tengMildi?
jЛ.
24) I IsWbifSthpX
-W
V.S'.S'.S’A
“AW.NSV.V
SS'.VA
W.:*v
P
4W*:¥:>-.sv;
• ••••г-х x * * 4 * * *
If - 0 tenglamaning ildizlaridan biri 2 ga teng.
teng.
36 = 0 tenglamaning ildizlaridan biri 3 ga teng.
nimaga teng.
26) Oshbii x2 + px + 45 = 0 tenglamaning ildizlaridan biri 5 ga teng.
P '.(-27) nimaga teng.
27) a ning qanday qiymatida x2 - (a-1)* + 36 = 0 tenglamaning
ildizlaridan biri 4 ga teng bo’ladi?
28) a ning qanday qiymatida x2 - {a + 3)* + 28 = 0 tenglamaning
ildizlaridan biri 7 ga teng bo’ladi?
SOLIYEV X. X.
#12#
01/11/2015 у
ккккк V KITOB kkkkk
29) a ning qanday qiymatida x2 - (2a - l)x + 56 = 0 tenglamaning
ildizlaridan biri 8 ga teng bo’ladi?
30) a ning qanday qiymatida x2 + (3a - 5)x + 72 = 0 tenglamaning
ildizlaridan biri 9 ga teng bo’ladi?
31) Ushbu 2x2 + x - a = 0 tenglamaning ildizlaridan biri 2 ga teng.
Ikkinchi ildizining qiymatini toping.
32) Ushbu x2 + 5x - 3a = 0 tenglamaning ildizlaridan biri 1 ga teng.
V.vly
Ikkinchi ildizining qiymatini toping.
33) Ushbu Ъх2 - 2x + a = 0 tenglamaning ildizlaridan biri 4 ga teng.
Ikkinchi ildizining qiymatini toping.
34) к ning qanday qiymatlarida kx2 - 6kx + 2k + 3 = 0 tenglama
kublarining yig’indisi 72 ga teng bo’ladi?
Xvlvlv.
*’ ‘' v<5;
35) Agar x - x + q = 0 tenglamaning val||f
.-л\\ч.
•fXsv
xl = 19
shartni qanoatlantirsa, q ning qiymati qanchaga tenglbo’ladi?
36) Agar x2 -x + q = 0
Ava ildizlari *,3 + x*
= 28
shartni qanoatlantirsa, q ning qiymati qanchaga teng bo’ladi?
.-X-X-x’x’x
37) Agar x2 -x + q = 0 tenglqmaning va; *2 ildizlari x* + x\ = 43
shartni qanoatlantirsa, q niislig^fl dianchaga teng bo’ladi?
38) *2 + px- 35 = 0 tenglamaning ildizlaridan biri 7 ga teng. Ikkinchi
•.N'jJA'.V.V.V.V.V.V.NVj.y.V.
ildizining va p ningf^iyinatihi toping.
39) x2 - px + 26 = 0 tehglamaijhg ildizlaridan biri 2 ga teng. Ikkinchi
ildizining va Jilting qiymatini toping.
40) * + цх -18Щ| tenglamaning ildizlaridan biri 3 ga teng. Ikkinchi
*“ ** * ’•’v!v!v!vXvlv!v,i \,!,Xv',|.vX\v.v.< \v!y‘
vi|p|pin|||iymatini toping.
I Xylylv/Xi 'yXvIvXvIvI vsy
44ШС Ills? 2.x-b tenglik * ning qanday qiymatlarida to’g’ri bo’ladi?
Jfc W 2
42) -V, va Щ sonlar x +x + a = 0 tenglamaning ildizlari bo’lib,
1 '•W.V.V.Vjb" N,v*s
= J tenglikni qanoatlantiradi. a ni toping.
43) va x2 sonlar x2 + x + a = 0 tenglamaning ildizlari bo’lib,
1 1
— + —
x, x.
1
- “7 tenglikni qanoatlantiradi. a ni toping.
1 л2 '
44) va x2 sonlar jc2 + x + a = 0 tenglamaning ildizlari bo’lib,
_ 1
- j tenglikni qanoatlantiradi. a ni toping.
1 1
SOLIYEV X. X.
#13#
01/1V2015 у
***** у KITOB *****
45)
*, va
*2
1
1
1
1 =
—
*i
*2
3
46)
*, va
*2
1 1
va
—+ —
“ —
*
2
47)
*, va
*2
1
1
1
*1
*2
3
48)
*, va
*2
1
1
1
— “b — —
- У
tenglikni qanoatlantiradi. a ni toping.
1
2 bo’lsa, a ning qiymatini toping.
47)*, va *2 sonlar x2 + ax-18 = 0 kvadrat tenglamaning yg§himlari va
2 bo’lsa, я ning qiymatini toping
-w
Jil
-w
ail
rife.
:.v.va
111
49) Ushbu x2 + px + 6 = 0 tenglama ildizlari ayirmaitpfig kvadrati 40 ga
teng bo’lsa, ildizlarining yig’indisin|janl|ia bo’lishini toping.
.v.vXwl^X'Iv.sv!’.'.
a s\;.v.v.,.,.v.,.nv.'y.v.% ,
50) Ushbu * + px + 4 = 0 tenglama ildizlari ayirmasining kvadrati 20 ga
" \;Xyv--'*X-Xy XyXyXyXyXyXv
teng bo’lsa, ildizlarining yig’indisini qancha bo’lishini toping.
51) Ushbu x2 + px-5 = 0/&nglal|a ildizlari ayirmasining kvadrati 84 ga
teng bo’lsa, ildizlarining yig’indisimidancha bo’lishini toping.
^ •:*:;:::::::::v:::-::v::;Xv;:;X:X;X:.... X£X*
52) Ushbu * + = 0 tenglama ildizlari ayirmasining kvadrati 40 ga
teng. pning
53) m ning qanday qiymatlarida 4*2 - (л/Зя? - 3)* - 9 = 0 tenglamaning
ildizlari qarama-qarshi sonlar bo’ladi?
.*" *XvIvIvXvXv.X X*X*’ .
54) m ning qanday qiymatlarida *2 - (V2/w - 6)x-3 = 0 tenglamaning
ildizl^b|lmmarqarshi sonlar bo’ladi?
55) m ning qanday qiymatlarida З*2 - (V5m - 10)*-4 = 0 tenglama-
\v!\^\;l4\\*X4\4\\*XyX\\*XvvX.
ning ildizlari qarama-qarshi sonlar bo’ladi?
***"**' О А ршшш
56) Rising qanday qiymatlarida 3* + (3m -15)* - 27 = 0 tenglama-ning
ildizlari qarama-qarshi sonlar bo’ladi?
57) m ning qanday qiymatlarida *2 + (2m -\2)x-\9 = 0 tenglamaning
ildizlari qarama-qarshi sonlar bo’ladi?
58) m ning qanday qiymatlarida 2*2 - (m - l)x -15 = 0 tenglamaning
ildizlari qarama-qarshi sonlar bo’ladi?
SOLI YE V X. X.
#14#
01/11/2015 у
***** у KITOB *****
2 о
59) к ningqanday qiymatlarida * +(k -4k-5)x + k = 0 tenglama-ning
ildizlari o’zaro qarama-qarshi bo’ladi?
60) к ning qanday qiymatlarida *2 + (k2 -7k-8)x-k = 0 tenglama-ning
ildizlari o’zaro qarama-qarshi bo’ladi?
61) к ning qanday qiymatlarida *2 + (k2 + 5k + 6)* + к = 0 tenglama-ning
ildizlari o’zaro qarama-qarshi bo’ladi?
2 2
62) к ning qanday qiymatlarida x +(k -7k + \2)x-k = 0 tenglama-
ning ildizlari o’zaro qarama-qarshi bo’ladi?
63) к ning qanday qiymatlarida x2 +{k2 - к-\2)x +к =^H|
ildizlari o’zaro qarama-qarshi bo’ladi?
64) к ning qanday qiymatlarida *2 + (k2 -2к-8)x-k = 0 tenglama-ning
ildizlari o’zaro qarama-qarshi bo’ladi?
65) к ning qanday qiymatlarida x2 +{k2 + 31I^MBM^P^ng1ama-ning
ildizlari o’zaro qarama-qarshi bo’ladi? Jt
ж
YjAWA'A . • V
AV.V.V.V.V.S
УЛ'Л'ЛЧЧ'Л’Л'Л*. .
• V.V.\\NV.'.4V.,.'.'.S*.',VS,AS4V
XVAWAV.'.'.V.'.'.’.'.'.VAV
•ANW.V.V.Vi’.'.SWAV
S'lWXVVAV^AVAVSWAV .
N\SVAN:AV,y,,.VAV.V.VA*
66) x + px + q = 0 tenglamaning ildizlari * - 3* #2 = 0 tenglamaning
ildizlaridan 2 marta katta bo’ 1 sa*|||f ^ qiymatini toping.
67) x2 + px + q = 0 tenglamanihgdldj^lari xl- 4* + 3 = 0 tenglamaning
t.V.V V.V.VAj.y'y.y.; V * V.4
ildizlaridan 3 marta katta bo’ls#i;feqiymatini toping.
68) x2 + px + q = 0
s\s\,\v.v.v.y.
V.SV.'AW.VAV
4‘.V.V.V.V.V.V.
i x -x-6 = 0 tenglamaning
ildizlaridan 2 marta katta bo’lsa,/? + q qiymatini toping.
.’.•.‘.V.V.V.V -
‘.V.V.V.V.V.VA*.
, *л*л*л*.ул*л В
70)
л: +
WAW.VAVJiA
•Л'Л’ЛЧ'ЛЧЛЛ',
•A’.*. *.•.*.•.•.4 V
•АЧ'/ЛЛЛу
69) x* + px + q = 0 tenglamaning ildizlari x2 - 5x + 6 = 0 tenglamaning
ildizlaridan 4 marta katta bo’lsa, p + q qiymatini toping.
ildizlari *2-8*-9 = 0 tenglamaning
ildizlarijip 2l||ii|ta:fitta bo’lsa, p + q qiymatini toping.
7ff *2 + px +Я- Q tenglamaning ildizlari *2 - 6* + 8 = 0 tenglamaning
5 marta katta bo’lsa, p + q qiymatini toping.
’vi'//iv!vIv!v!vI,Iv!v!,Xv!,!v!,Iv.
72) f2 + px + q = 0 tenglamaning ildizlari *2 -*-12 = 0 tenglamaning
ildizlaridan 8 marta katta bo’lsa, p + q qiymatini toping.
2
73) Ushbu * - 5* + a = 0 tenglamaning ildizlaridan biri ikkinchisidan
9 marta katta bo’lsa, a ning qiymatini toping.
74) Ushbu *2 —10* + a = 0 tenglamaning ildizlaridan biri ikkinchi-sidan
4 marta katta bo’lsa, a ning qiymatini toping.
75) m ning qanday qiymatlarida *2 - 4mx + 48 = 0 tenglamaning
ildizlaridan biri boshqasidan 3 marta katta bo’ladi?
SOLI YEV X- X.
#15#
01/11/2015 у
***** у KITOB *****
76) m ning qanday qiymatlarida *2 - 5mx + 36 = 0 tenglamaning
ildizlaridan biri boshqasidan 4 marta katta bo’ladi?
77) m ning qanday qiymatlarida * - 2mx + 80 = 0 tenglamaning
ildizlaridan biri boshqasidan 6 marta katta bo’ladi?
л
78) m ning qanday qiymatlarida x - mx + 24 = 0 tenglamaning
ildizlaridan biri boshqasidan 6 marta katta bo’ladi?
A _
79) m ning qanday qiymatlarida * -3mx + 12 = Q tenglamaning
ildizlaridan biri boshqasidan 2 marta katta bo’ladi?
1
80) Ildizlari * + px + q = 0 tenglamaning ildizlariga teskgflqllgan
1Щ
tenglamani toping.
81) Ildizlari ax2 +bx + c = 0 tenglamaning ildizlariga teskari bo’lgan
tenglamani toping.
82) x2-0,5kx + k2-\\k + 24 = 0, (k =
tenglamaning ildizlaridan
biri 0 ga teng. Shu shartni qanoatlantiruychi ildizlarining yi’g’indisini
toping.
83) x2 - 3kx + k2 - 2k - 3 = 0, (k = tenglamaning ildizlaridan biri
0 ga teng. Shu shartni qanoatlaptiruvchi ii$2larining yi’g’indisini toping.
VAWA'A’A’A
ViVAJAWA'A
‘.V.VA’A’A*
' 4MvM*y
84) x2 - 5kx + k2 - Ik +12 ) tenglamaning ildizlaridan biri
0 ga teng. Shu shartni qandatlantiruvchi ildizlarining yi’g’indisini toping.
85) x -kx + k - 9к+ШШФ^. (к = bonst) tenglamaning ildizlaridan biri
‘•yXvXvlvlvlvXyvX^
0 ga teng. Shu shartn|panoatlahtiruvchi ildizlarining yi’g’indisini toping.
86) x2 -Ojlkxgk20, (k = const) tenglamaning ildizlaridan biri
0 ga teng. Shu shartni qdnoatlantiruvchi ildizlarining yi’g’indisini toping.
87) x
biriO
\,.V.SW.V.V.v.%
WAWAWA
'.•.V.V.V.VWA
A\VA'.4*.y.VA
S‘A
VHOOKas,.'.>.sn>i
• A'.V.V.V.V,*.
vftXvNv.4\v.*.v.
. .........
SWAWASV .
•
V.SW.V.V.;
v.VAaSH
wfoo бос
v.w*
12 = 0 , (к = const) tenglamaning ildizlaridan
i qanoatlantiruvchi ildizlarining yi’g’indisini
^ • sw.v.v.sw.v.*. v.\*
WSSW.SV.V/A'A .*AV
a vKVA,A4‘A'i|,\ _ . л
mgsmmk -13) - к= 0 tenglamaning ildizlari qarama-qarshi
sonlar bo’ladigan к ning barcha qiymatlari yig’indisini toping..
89) л;2W (3k2 -6k-\7)~k4 =0 tenglamaning ildizlari qarama-qarshi
sonlar bo’ladigan к ning barcha qiymatlari yig’indisini toping..
90) 3*2 +{6k2 — 9k + 2) — k4 = 0 tenglamaning ildizlari qarama-qarshi
sonlar bo’ladigan к ning barcha qiymatlari yig’indisini toping..
2 2 —
91) 5* +(/: -Ik-6)-к = 0 tenglamaning ildizlari qarama-qarshi
sonlar bo’ladigan к ning barcha qiymatlari yig’indisini toping..
SOLIYEV X. X.
#16#
01/11/2015 у
***** V KITOB *****
ill,
::¥й:й<:
92) 2x2 + (9k2 - \2k + 2) - k4 =0 tenglamaning ildizlari qarama-qarshi
sonlar bo’ladigan к ning barcha qiymatlari yig’indisini toping..
2 2
93) к ning qanday qiymatlarida (k-2)x +7x-2k =0 tenglama
x = 2 yechimga ega?
94) к ning qanday qiymatlarida kx2 + \2x - 3 = 0 tenglamaning
ildizlaridan biri 0,2 ga teng bo’ladi?
95) к ning qanday qiymatlarida x2 + kx- 5 = 0 tenglamaning ildizlaridan
biri 3 ga teng bo’ladi?
96) к ning qanday qiymatlarida kx2 - 4x +1 = 0 tenglamlmjp
ildizlaridan biri 0,5 ga teng bo’ladi?
97) к ning qanday qiymatlarida x2 +5x-7к = 0 tenglamaningjr
ildizlaridan biri 2 ga teng bo’ladi?
98) Ushbu x2 + 2ax + a = 0 tenglamaning ildizlSi^ birilllga teng.
Tenglamaning ikkinchi ildizini toping.
99) Ushbu x2 - 2ax + 3a = 0 tenglamamhg ildizlaridan biri 2 ga teng.
Tenglamaning ikkinchi ildizini toping.
,, Jm*.
4 b +a .
100) Agar - - 2 bo
nimaga teng bo’ladi.
.yjSSSSWJc* 'л*. •.
A4\SV.Vt,.Vi‘« .\S* •.V.VA'tV.V.V.V
4 b + a 2o21 Wab- 2
nimaga teng bp’ladi.llli,
.v.v.v.v.w
«Ik
УУУЛу
"W
mm
• v.v.
Ж
Ж
'Illl!::
2
Sk
ifodaning qiymati
-7 b
ifodaning qiymati
,;s TOW'Xv
* Of*'-
nimaga
ШШШ
: :
\\*XvKvX*Xv.*.
•*л:лчул\ул\
y.v.v.v.y.
a +6ab-5b
2a2 +9b2
ifodaning qiymati
rliir
Щ Щ||щЩ|ау musbat qiymatida 8*2 - 30* + a3 =0 tenglama-ning
ildizlaridan biri ikkinchisining kvadratiga teng bo’ladi?
V^!vyX;VXv!;’\vIvIvlvX£X*XvX л a
104) arning qanday musbat qiymatida * -6x + cr = 0 tenglama- ning
ildizlaridan biri ikkinchisining kvadratiga teng bo’ladi?
105) a ning qanday musbat qiymatida *2 - 2* + a3 = 0 tenglama- ning
ildizlaridan biri ikkinchisining kvadratiga teng bo’ladi?
106) va *2 sonlar 3*2 +2x + b = 0 tenglamaning ildizlari bo’lib,
2*, =-3x2 ekanligi ma’lum bo’lsa, b ning qiymatini toping.
107) к ning qanday qiymatida 5* - 22* + 5k = 0 tenglamaning
•*, va *2 ildizlari orasida 5*, + 6*2 =26 munosabat o’rinli bo’ladi?
SOLIYE V X. X.
#17#
01/11/2015 у
***** V KITOB *****
A
108) к ning qanday qiymatida 3* — 8л: + ЗЛ: = 0 tenglamaning
X va *2 ildizlari orasida 6x] +1 lx, = 31 munosabat o’rinli bo’ladi?
^ _
109) к ning qanday qiymatida Зд + 5д + ЗА: = 0 tenglamaning
X va x2 ildizlari orasida 6д, + 8д2 = -14 munosabat o’rinli bo’ladi?
110) A: ning qanday qiymatida 4x2 -7x + 4k = 0 tenglamaning
X va x2 ildizlari orasida 16.x, + 3д2 = 2 munosabat o’rinli bo’ladi?
111) к ning qanday qiymatida Зд:2 — 10л: -+- ЗА: = 0 tenglamaimg
X va x ildizlari orasida 6x + 7д, =24 munosabat o’rinli bo’ladi?
112) к ning qanday qiymatida 5x -\2x + 5k = 0 tenglamaning
\s\44sVi.. л,лул,лччу.улч,л,л.
шМшт
X va x2 ildizlari orasida 5д, + 8д2 =21 munosabat o’rinli bo’
^ ....
113) к ning qanday qiymatida Зд -1 Од + ЗА: tenglamaning
.•X*. w
лул л;. .
X va x2 ildizlari orasida 6д, + 13д2 =48 munosabat o’rinli bo’ladi?
I Л* I 4* •Xs\s\\-XvXvX\*X\*XvX«X%-.\-.. *vX*X‘X-‘
• ',NV.SVS\VA\S\yAWA4yA,A\ . WiV
ж
114) к ning qanday qiymatida 4д -17x + 4A: = 0 tenglamaning
X va x2 ildizlari orasida 8д, + 11д2 =J9 munosabat o’rinli bo’ladi?
i I • ,'X»Xy. vy,
-■ШШЬу- % ЧР
115) A: ning qanday qiymatida lx ч-;:1 8д +C7k = 0 tenglamaning
X va x ildizlari orasida 14x -9 munosabat o’rinli bo’ladi?
A'.v.v.v.v.s'.s л'.у. •••!•!••
^y.vy.y.y.y.'.'.y^ A \\\V\ sy
116) A: ning qanday qiym^|jda^JP^%s3A: = 0 tenglamaning
•й%'й*л
X va x ildizlari orasidat6xl+ 9xllMMiunosabat o’rinli bo’ladi?
117) Agar a, b Gj||a (a-b)2 =10 bo’Isa, a + b yig’indi qanday
qiymatlami qabul qil||||,
1
. \S4WA\4N\\V.V. .
•лулчулчул
118) Ашж, В111Ж vtifif-b)2 = 8 bo’lsa, a + b yig’indi qanday
. xW>x:>x:x:::x :::;x:x
Я№ШЖ
V.4\S,.V.\\V.y.
ул-л'.у.у.ул-
•ГА
1
N va *b)
qiladi.
2
= 12 bo’lsa, a + b yig’indi qanday
Ч о •% • •
.v.v
120) Agar7/, b e N va. (а-b)2 =15 bo’lsa, a + b yig’indi qanday
qiymatlami qabul qiladi.
121) Agar x2 -4ax + la2 =0 tenglamaning ildizlari x va x2 lar
uchun x2 + x2 =2 tenglik o’rinli bo’lsa, a2 ning qiymatini toping.
122) x2 -4x-(a- 1)(я-5) = 0 tenglamaning ildizlaridan biri 2 ga teng
bo’ladigan a ning barcha qiymatlarini toping.
#18#
01/11/2015 у
***** у к| jo в *****
123) q ning qanday qiymatlarida x2 - 8x + q = 0 tenglamaning
ildizlaridan biri boshqasidan 3 marta katta bo’ladi?
124) 5x2 +bx-2& = 0 tenglamaning ildizlari va x2 laruchun
5x{ + 2x2 = \ munosabat o’rinli. Agar b butun son ekanligi ma’lum bo’Isa,
uning qiymatini toping.
125) x2 — {a +14)* + a2 = 0 (a> 0) tenglamaning ildizlari orasida
= 9x2 munosabat o’rinli. Berilgan tenglamaning katta ildizini toping.
2
126) a ning qanday qiymatlarida * + 2(1 - a)x + a + 5 = Qlleiiglama- gning
yechimlari o’zaro teng bo’ladi?
127) a ning qanday qiymatlarida cix2 - {a + l)x -t- 2л — 1 = 0 tenglffia bitta
ildizga ega bo’ladi? j| ™
_ Л *AV
128) к ning qanday musbat qiymatida 25x :+kx + 2 = Oitenglama bitta
ildizga ega bo’ladi?
129) a ning qanday qiymatlarida сГхШ+ 2x +1 = (|lenglama bitta ildizga
ega bo’ladi?
wav.v\4\\v.
.v.4WA'ASV.V.V.v.v.ViN> , .
* * WA,ASSW.N%V.SVA4W>A,?<.
• n%n;.n4V^v.;.w\v.v.s\v.v.\;ii e
%
•:x:x‘
130) (xx2 =|c/| tenglama yagona^ydqjnmgaega bo’ladigan a ning barcha
Jltlk “
r.V.-.V%.VA-.V.\
Л* V,\V.-.V.«.-.N\
V.V.VA'tV.V.Wk
шщ+2
+1 )-k =0 tenglamanoldan
qiymatlarini toping.
131) к ning qanday qiyiptt|andi||lr
farqli o’zaro teng ildizlarga ega?
132) Ushbu y2 - 20k t + 2 чЙ tenglama faqat bitta ildizga ega
wXjIv!j^sn\\ ;•
bo’ladigan t ning barcha qiymalari yig’indisini toping.
133) a ning qanday qiymatlarida ax2 - 2x + 3 = 0 tenglama bitta ildizga
ega bo’ladgL '•k
134) /7||pingqanday qiymatlarida x +(m + 2)x + m +5 = 0 tenglama-
ningijdizian haqiqiy va o’zaro teng bo’ladi?
* * wli .w.v
/.J.y " A ,.V.V,y.V«,.*.\\V.i y.S| «
x-к+ 6 tenglamaning ildizlari bir-birga teng
ЬоЩ|р1р ning barcha qiymatlari ko’paytmasini toping.
A
136) a ning nechta qiymatida
ega?
A
137) m ning qanday qiymatlarida (w - 2)x - 2mx + 2m-2> = 0 tengla-ma
bitta ildizga ega bo’ladigan qiymatlarining o’rta arifmetigini toping.
A
138) к ning qanday qiymatlarida kx -{k-l)x+ 9 = 0 tenglama ikkita
teng manfiy ildizga ega?
3x-a x+ a
1
3-x x+1
= 2
tenglama bitta ildizga
X.X.
#19#
01/1V2015 у
***** у KITOB *****
2
139) a ning qanday qiymatlarida * - {a - 2)x - a -1 = 0 tenglama
ildizlari kvadratlari yig’indisi eng kichik qiymatga ega bo’ladi?
140) m ning qanday qiymatida x2 +(m +1)* + m2 -1,5 = 0 tenglama
ildizlari kvadratlarining yig’indisi eng katta bo’ladi?
141) m ning qanday qiymatida x2 + (2- m)x - m - 3 = 0 tenglama
ildizlari kvadratlarining yig’indisi eng kichik bo’ladi?
2
142) a ning qanday qiymatida * + (a + 2)x + a = 3 tenglanm ildizlari
kvadratlarining yig’indisi eng kichik bo’ladi?
143) va *2 x2 -px + p-1 = 0 tenglamaning ildizlari:^l^l|||pndif
qiymatida x2 + x2 yig’indi eng kichik qiymatni qabul qiladi?
144) y] va y2 y2-by + b-1 = 0 tenglamaningJldizlari bo’Isa, b ning
qanday qiymatida y\ + y\ ifodaning qiymati eng kichik bo’ladi?
145) q ning qanday qiymatida x2 - %x + q = 0 tenglama ildizlari
kvadratlarining yig’indisi 34 ga teng bo’ladi?
. -<v
W/.V.V.V.V. J,|IV
146) m ning qanday qiymatida 3;й11Ш:У:+ m- 0 tenglama ildizlari
kvadratlarining yig’indisi 25 ga teng bo’ladi?
.V.V.V.'/.V.N
‘.V.V.V.VAW
VAWAW
147) a ning qanday aiутаШШШЛш^- 2)jf + a uchhad ildizlari
kvadratlarining yig’indisi 3 ga tengbg’ladi?
148) a parametrning qand4^butun qiymatida 2x2 + 6ax + a = 0
tenglama ildizlari kvadratlarining yig’indisi 38 ga teng bo’ladi?
149) a ning qanday qiymatlarida son o’qida x2 + ax +12 = 0
tenglamaning ildizlari orasidagi nasofa 1 ga teng bo’ladi?
\\\v.s%v»s\s%v •
.Ящк
150) Ч|!М №= 0 tenglamaning yechimlari xl va x2 bo’Isa,
\xi - x2[Щ. mundsattll p ning nechta qiymatida bajariladi?
n riing qanday qiymatlarida x2 -12x + n = 0 tenglama ildizlari-dan
bin ikkmchisidan 2V5 ga ortiq bo’ladi?
-.чу.чЧЧ-- 2
152) ning qanday qiymatida * -px + 5 = 0 tenglamaning ildizlari biri
boshqasidan 4 ga katta?
4V.V.V.VAV!\4
2
153) m ning qanday qiymatlarida 4x - (3 + 2m)x + 2 = 0 tenglama-ning
ildizlari biri ikkinchisidan 8 marta kichik bo’ladi?
154) a ning qanday qiymatida 4*2 — 15л: -f- 4<^2 = 0 tenglamaning
ildizlari biri ikkinchi ildizining kvadratiga teng bo’ladi?
155) Agar x2 - 3x + m = 0 tenglamaning xl va x2 ildizlari uchun
Здг, - 2x2 = 14 munosabat o’rinli bo’Isa, m ning qiymatini toping.
SOLIYE V X. X.
#20#
01/11/2015 у
***** у KITOB *****
156) х, va х2 sonlar x2 + Зх + к = 0 tenglamaning ildizlari
bo’Isa, к ning qiymatini toping.
A
157) * + x + a = 0 tenglamaning x, va x2 ildizlari orasida
i _
2
5
1 1
—+ —
*1 *2
2
*r
is
• V.V.VA%%V.V.W.,iVAV.%V.W/
• • V»> V» V# V •: •
munosabat o’rinli bo’lsa, a ning qiymatini toping.
158) x2 - (a + 2)x + « + 7 = 0 tenglama ildizlariga teskari soglar
7
yig’indisi ~ gateng bo’lsa, a ni toping.
159) 2л2 - 7x + c = 0 tenglamaning ildizlaridan biri 0,5 ga teng.
tenglamaning ikkinchi ildzini toping.
160) q ning qanday qiymatida x2 - x - q = Q tenglama ildizlari kublari
yig’indisi 19 ga teng bo’ladi?
161) m ning qanday qiymatlarida (m if|)x2 + 2тх£ЩР- 2 kvadrat
uchhadni to’la kvadrat shaklida tasvtriasbmumkin?
162) x, va x2 lar x2 + |«|x + 6 = J||eliiapidning ildizlari bo’lib,
x,2 + x2 = 13 tenglikni qanoatlantirsa, x1 + x2 nechaga teng?
^ ''ivlvlvlyXvXv., vXv
163) ax +bx + c = 0 ten|lamaninf:koeffisientlari b = a + c tenglikni
'щшш.
qanoatlantiradi. Agar x, va x2 berilgan kvadrat tenglamaning ildizlari
.•.V.V/.’A*
VAW
vXv/’.v
fillip
bo’lsa, +
*1
1
*2.:&
I\4W»W>.VAw
A'.v.v.v.’.vi*.
.•.■.'.■/.V.V.V.V.V.
• •••
•AWAW. 4*.SV.*.'.
VA'.y.V.S'.VA'A .
s|rp:«#
i'.v
•.v
VM«
164) ax + Щ§||= 0 tenglamaning koeffisientlari b = a + c tenglikni
VB®»yw»\sy.v ...
/.v.y.v.. t \j.v.\\v.va;.4v .
qanoatlantiradi. Agq|i;:$lva x2 berilgan kvadrat tenglamaning ildizlari
ill*,, *llllb
N'AWA’AWAV
swV.v.v.y.*
VAW.VAV
•\-;y
Ж
ЩШШЖ О Л
ning qiymatini hisoblang.
a + b
Agar 2a +2b = Sab va 6 > « > 0 bo’lsa, q_^ kasming qiymati
*.*.*.’.V.N . . .Л
'.■.V.V.V.VA’.SN
•.•Л’Л'А'Л’Л'Л'
166) x2 - 3«x + la2 -ab-b2 = 0 tenglamani yeching.
л
167) S - kvadratning yuzi x -Sx + 9 = 0 tenglama hech bo’lmaganda
bitta ildizga ega bo’lishi uchun kvadratning a tomoni qanday bo’lishi
kerak?
#21#
Ol/U/2015 у
***** у KITOB *****
168) 2 va -3 sonlari x3 + mx + n ko’phadning ildizlari. Bu ko’phadning
uchinchi ildizini toping.
169) kx2 + 3kx + 2k-\ = 0 tenglama yechimga ega bo’ladigan к ning
butun qiymatlari o’rta arifmetigini toping.
170) ^ ning qanday qiymatlarida
ildizga ega?
171) a ning qanday qiymatlarida
, a
x + 4 = —
tenglama ikkita turli haqiqiy
т a
x-3 = —
‘.VA'.V.y
'•X'Xv/XvX
.V.V.*
•vXvXv!
;.>v.
v.v.v
*******
•Xv
IvXv
::
m
ildizga ega?
172) a ning qanday qiymatlarida 5{a + 4)x2 - Юл + a = 0 tengl|ml-
ning ildizlari turli ishorali bo’ladi?
173) a ning qanday qiymatlarida xl + Зл: +JBI||& = 0 tinglamaning
ikkala ildizi ham manfiy bo’ladi?
174) p ning qanday qiymatlarida x~gf%p + 1)л: + *gp - 5 = 0 tenglama-
ning ikkala ildizi manfiy va turli bo’
175) к ning 4у -3y + k = 0 tenglama haqiqiy ildizlarga ega
bo’lmaydigan eng kichik qiymattni toping.
176) Agar m va n sonlar x2 + 3mx^5n = 0 (/я • и # 0) tenglamaning
ildizlari bo’Isa, n-m m^qiymati nibhaga teng bo’ladi?
177) va у2 У + ту + n =^|englamaning ildizlari y{ va y2 ning har
birini 4 taga orttirib, il^faari hosil bo’lgan sonlarga teng bo’lgan kvadrat
tenglama tuzifflllAgar umlfozod hadi n-24(n dastlabki tenglamaning
ozod hadi) ga teng bo’Isa, m nechaga teng?
178) Ushbu:;i^^p^|* <7 = 0 tenglamaning har bir ildizini 4 taga orttirib,
ildizlari ho|il Bo’lgan sonlarga teng bo’lgan kvadrat tenglama tuzildi.
Agar uning ozod hadi <7 + 64 ga teng bo’Isa, p nechaga teng bo’ladi?
2ax + a2 -1 = 0 tenglamaning ikkala ildizi - 2 va 4 orasida
joylashgan bo’Isa, a ning qiymati qaysi oraliqda o’zgaradi?
180) x2 - 2ax + a2 - 4 = 0 tenglamaning ikkala ildizi - 4 va 7 orasida
joylashgan bo’Isa, a ning qiymati qaysi oraliqda o’zgaradi?
181) n natural son n2x2 + 3n3x + 4 = 0 tenglama ildizlarining o’rta
arifmetigini o’rta geometrigiga nisbati -3 ga teng bo’lsa, n ning
qiymatini toping.
182) - la2x2 - 9a4 =0 (а ф 0) tenglamaning haqiqiy ildizlari nechta?
183) *4 - 3a2x2 - 4aA =0 (а Ф 0) tenglamaning haqiqiy ildizlari nechta?
x
#22#
01/11/2015 у
***** VKITOB *****
1. Parametrli tengsizliklarni yeching.
1) Tengsizliklar sistemasi a ning qanday qiymatlarida yechimga ega
bo’lmaydi?
ax >5<7-l fax >7a-3
2) 3)
i)
4)
ax < 3a+ 1 ’
3-lx <3x-l
\ + 2x < a + x
5)
ax < 3a + 3 ’
2x-3> x+ 2
4 + 5x <<7 + 4* ’
ox > 9a - 5
^£ 6o+7;, Д
6x-ll>13-2x
6> 7.v - 5 < « +
ШШ
w
O.V.^V.SW.VAW.V.VA'A,
1.
A’.'.W.V.VNS .
.•.•.•.•«V.V.VAV.V. .
f -.V-V-V-WS*,'
•AV.'/.'.SSS’.V.N'.V.V.'.'.V/.NV.VA
' v\V.\\V.V.4V.44V.V.\V.*.'
.*X\
6) a ning qanday qiymlgarida
2) Tengsizliklar sistemasi b ning qanday qiymatlarida yechimgplga
bo’lmaydi?
bx> 6b-2 fbx> 5b-3
^ \bx<4a + 2 ; 2) [Ax<4A + 3 ; j|3^ [Ax<2g4f*
3) a ning qanday qiymatlarida a(xj||j)fe i - 2 tengsizlik x ning barcha
qiymatlarida o’rinli bo’ladi?
4) a ning qanday qiymatlaridii?(3x- 5) > 6x-13 tengsizlik x ning
barcha qiymatlarida o’rinli bo’laffll
5) a ning qanday qiymatlarida арзЦЦ < 8x + 9 tengsizlik x ning barcha
qiymatlarida o’rinli::|q’illi||
x-l)>21x-8 tengsizlik x ning
barcha qiymatilrida o’rinli bo’ladi?
\sy.\v.\v.vy.^;.v _ vwA'.y.y.sV ^ _
7) Ushbu (* - Щх-b) < 0 tengsizlikning yechimlari [2 ; 6] oraliqdan
V v^wawaw .sw.*
iborat. |J; toping.
8) ШЬЙЯЬ- Щх -Ъ) < 0 tengsizlikning yechimlari fl ; 9] oraliqdan
yly/XyXyXyly vXv'v
iborat. я+A ning qiymatinitoping.
X\<\\vav/X‘a\<%444\v\4!s^ . * "vXyivlv'-' - . » « «
9) Ushbu (x - a\x - A) < 0 tengsizlikning yechimlari [— 3 ; 5J oraliqdan
iborat|feg5|i ning qiymatini toping.
10) Ushbu (x-aX*-A)<0 tengsizlikning yechimlari [-7 ; 3] oraliqdan
iborat. a + b ning qiymatini toping.
A
11) Ax + 2x+& + 2>0 tengsizlik yechimga ega bo’lmaydigan к
ning butun qiymatlari oraisda eng kattasini toping.
12) a ning qanday qiymatlarida ax2 + 8x + a < 0 tengsizlik x ning
barcha qiymatlarida o’rinli bo’ladi?
SOLIYEV X. X.
#23#
01/11/2015 у
***** у kitob *****
A
13) A ning kx + 4л: ч- Л: +1 > 0 tengsizlik yechimga ega bo’lmaydigan
butun qiymatlari orasida eng kattasini toping.
14) t ningqanday qiymatlarida tx2 -6x-1<0 tengsizlik x ning barcha
qiymatlarida o’rinli bo’ladi?
15) a ning qanday qiymatlarida (a -\)x2 - (a + \)x + (a + \) <0 tengsizlik
x ning barcha haqiqiy qiymatlarida o’rinli bo’ladi?
16) a ning qanday qiymatlarida (2 - a)x2 + 2(3 - 2a)x - 5a + 6 < 0
MW
tengsizlik x ning barcha haqiqiy qiymatlarida o’rinli bo’ladi?
17) n ning 10 dan oshmaydigan nechta natural qiymatida
nx2 + Ax > 1 -3/2 tengsizlik л ning ixtiyoriy qiymatida о’пп1ГЩ1||р?
mx + 9
18) m ning qanday qiymatida --iy tengsiziiKning eng
л;.ул\,.,л
Щ0
>-10
.v.w.. .-л-
AV.V.VA'.V.VAWA .
v.y.vy * -
... . .VAWAWA'AV.wX’AN
• Ч’.'.'Л’Л'Л'Л’ЛУ.'Л'Л'Л'Л’Л'.'. •
* wavavawnvmSmmSS®
'.,Л,Л,Л'Л,Л,Л
• v.SSSW.y.
manfiy yechimi -3 ga teng bo’ladi?
nix + 5 q
19) m ning qanday qiymatida ' - ■ tengsizlip^ng eng katta
ЛШ'-
manfiy yechimi -5 ga teng bo’ladi?!!!;^.
mr
w
20) Tengsizlikni yeching. (а; я* ->Щ§
••••••••• •.V.VA'.V.VAW. V ,'AV
.V.... '.yA’.'.W.V.S'.V. VAV
Ши ЩШжк. m i
21) Tengsizlikni yechi|§y^> o) ||jJp:::> ~
“■111:.
|\SWH
A\’.V.VA
w
TKiXSIZLI IiLAltM ISBOTLASH.
ШЖь
4»
1
‘•a'A'XvI'aXv'a
•A'.SV.SSV.V.
Cv
a >
|&| bo’Isa,
1
1
a3 +b3 ’ a4 +b
1
a
larni
2) Ushbu fengsizliklarning qaysi biri a ning barcha qiymatlarida o’rinli?
1 2
*v::Av:vA'>4V
1. a2>0 ; 2. o2-10<0 ; 3. (a-5)2>0;
^2 j ^
5. a2 + 7 >0 ; 6)(2я + 3)2<0; 7)
1 1 1
4. —
a
+ a* >2 ■
<0
3) a>b> c> 0 bo’lsa,
—, 7 va —;—
a a+b a+c
larni taqqoslang.
#24#
01/1V2015 у
***** у ютов *****
1 1 1
г va lami taqqoslang.
a+b a-c ^ b
a
4) a <b <c< 0 bo’lsa,
5) a>b> 0 shartni qanoatlantiruvchi va b sonlar uchun quyidagi
munosabar qaysilari o’rinli bo’ladi?
2r 2
1. a3 >ab2 ; 2. a4>azb
2 2
3. a2b2 < bA ; 4.->T.
’ <7 о
#i
6) Agar a, ieiV, я > 10, b> 16 bo’lsa, quyidagilardan qaysi biri har
doim o’rinli bo’ladi?
3 a-b л 6-2я _ Мш2:Ь
1. a-i<0; 2. —r“>0; 3.
b a
7) Agar * va у sonlari uchun *-y = 20 va 0<л:<0,8 munosabat <f rinli
bo’lsa, quyidagi tengsizliklardan qaysi biri doim| o’rinli bo’ladi?
<0-
silk
1 -<2°; 2. * + y < 20 ; 3. У <16 ; 4«р||
8) Agar 2 <a<3 va -3<b <-2 bo’lsa, quyidagilarni qaysi biri har
doim o’rinli bo’ladi?
v%vXv//Xv.w/Iv.v.v. *«V
V-.
l.aV -50 < 0
5 < 0
4.Л2-2<0 ;
\p2+q2<m
Э. Agar |
тшт чч
5. W
.•XyX
WiSSX'.v
butun qiymatlari nechta?
10. Agar - 2 ?g|r < -1 ®;k 3 <4S < -2,5 bo’lsa, я - 6 ayirma qaysi sonlar
oraisda bo’ladi? : : %Jp|lr
11. AggP^fcWlll§#quiydagi keltirilgan ifodalardan qaysi birining
^л;лу 4X*Avlv!'X¥A‘.-.
qiymati
4. a
5. a
1 .IPfKp + Vl997 va /> = 2 Л996 ;
2. а ЩШб + 4\Ш va = 2vff;W ;
3. с = л/Гз -t/12 va ^ = Vl4 - Vl3 ;
4. o = ЛоТ + л/ШЗ, 6 = V99 + Vl05 , c = 19,9
5. с = Л2 + Л5 , </ = ЛТ + Л7
6. m = \l3, n = V2, р = Ло
7. a = 1-2-3-4-...-29 va 6 = 1529
-/ж* у
#25#
Ol/H/2015 у
***** у KITOB *****
59
8. a = 1-2*3-459 va b-30
13. Quyidagi tengsizliklardan qaysi biri * va у ning xy> 0 shartni
qanoatlantiradigan barcha qiymatlarda o’rinli?
1. (л — yf > 0 ; 2. —+ — S 2
4 7 ’ у д:
4. ДГ -6лу + 9у2 < 0 ; 5. л:3 — >»3 > 0 .
14. Quyidagi munosabatlardan qaysi biri noto’g’ri?
3. *2 -y2 >0
1. 2.
4.
a = a
5.
a5 +b
a2 +b
>a5+b
<a2+b
3.
a3 +b
> a3jg£%;
.ОлУ'/л
ill
w
15. a>2b>0 shartni qanoatlantiruvchi a va b sonlar uchun quyidagi
munosabatlardan qaysilari o’rinli?
1. a3 >7b
a-b b
2 > -
2 2
6b- а л
3. <2
a
16. Agar ci<b va ab # 0 bo’lsa, quyidagi tengsizliklardan qaysi biri
har doim o’rinli bo’ladi?
* V.W.V.V.V.VAV
\v.v.y.y.;.;.svA
3. -тШь-h.: 4/lP< 3a+ b : 5. 2a > a + b.
6b Щш л
Ipi»— > 0
•%4\
wOQQW
a
Jl
1 1 , ,
1. ->T ; 2. o2 >62 ;
a b ' ’
17. Agar * > у va 2 > t bo’lsa, quiydagi tengsizliklardan qaysi biri har
doim o’rinli bo’ladi?
M чт®щт*>. V
1. *• г > уt; Ill > jRfl|3.
4. It -13* < 7jr-13ДГ-1!' > y-t .
X:XvXv>X:A;>.
ж
(z + /V
1
V.v/.y.v.y.v.vA
18. Agar — < ^«bo’lsa, quyidagi ifodalardan qaysi birining qiymati eng
Ш
у/.у.улу.у.у
4vXv:::v'
'w* / <й::.ж.й.5й¥х
ЯЁ ... ...
SV.4V . .■.■.•.•Л'Л*. . •.V.V.V.V.V.-.V.V.W
SVAS\. . .V.V.V.'/A'.'.'.S
SVVAWAVA^^^WAW.V'V. * V.V/.V?.V *
•VASyAV» A’A'iV.SV/iS'.SWiV.Vi
3. a3 -1 ; 4. я2 -1 ; 5. 1-tf
1
19. Agar Щ > у, t - - bo’lsa, quiydagilardan qaysi biri doimo o’rinli
bo’lall^'
111 1 1
i t + — = z + — ■ 2 x + -< y + z ■ з x + -> y + z ■ 4 дм— > V •
л: У ’ / ’ t ’ z ’
1 1
5. *+->y+- .
/ 2
20. Agar * > 3 va у < -3 bo’lsa, quyidagi tengsizliklardan qaysi doimo
o’rinli bo’ladi?
SOLIYE V X. X.
#26#
03/11/2015 у
***** V KITOB *****
1. (^ + ^)2 >3 ; 2. x-y<-9 ; 3. ->-! ; 4. >6 ;
jc -v 1
5. 2 2" .
* +_y 2
ly 1 19 I y2
21. Agar xe[2 ; 5] va ye[2; 5] bo’lsa. +jh + 1
ifodaning eng katta qiymati nechaga teng bo’lishi mukin?
22. Agar 3<x<6 va 15 < < 60 bo’lsa, ~ ningqi
X
tegishli?
23. Agar 2 < д < 5 va 3 < v < 6 bo’lsa, xy-x mng qiymati qaysi
oraliqqa tegishli bo’ladi?
Jk3
sO'XvI'.XwXvI'.v.
•*•••.......
Xviy. .•
.v.y/.y.v
•УУ.У.У
.m
•v.v.v.uw/.v.vXvX
24. Agar 16 < д < у < z < t < 100 bo’
X
.NV.VX
,\v.v*v.s
AW.V.S*.'
•WAV.W.
V.W.V.S4
:;w
qiymatini toping.
tVA
..Si
.v.»y.\
'.S*v
Л'Л'Л’Л
.y /
eng kichik
•ii;,
Ш'
25. Agar 9 < x < y< z <t < 81 bo’lsa, ^#11:,ifodaning eng kichik
qiymatini toping.
•ч\
A‘A
MV
AW.V.
v.v.v.s
%
у
V.VAV.WAV.V
VAWMVAV ORB
v v
26. Agar 25<*<Х<ЩШ#4 bo’lsa.
qiymatini toping.
^ t
ifodaning eng kichik
v.v.v.v.v.,.v.*.
Ilf
27.
о».
jmKbXv.sv '
"111Ы
Agar 7 < x<y<z< t<\\2 bo’lsa, ~ + ~ ifodaning eng kichik
I» У 1
• Д • 'VAV/JlvIyilvX ViVAVAViVA*.1 .VAV
fOM2. “ -
ft.
:Ш?й.
йч
Л’Л
- .‘ASV.S
VAVAV.V.V
VA'AV
: ‘
AgaP|li.xf ^ < z < / < 121 bo’lsa,
—+ —
^ *
ifodaning eng kichik
• w.v.v.
XV
AVA .
VAWA • . .
A'/A'A'A'
Щ
29. Aglf 8 <x<y<z<t< 200 bo’lsa,
— + —
У 1
ifodaning eng kichik
• •
qiymatini toping.
30. Agar 5 <x<y<z<t< 320 bo’lsa, ~ + ~ ifodaning eng kichik
qiymatini toping.
лшм л
X.
#27#
01/11/2015 у
***** у к|"ГС)В *****
31.
i
——— ifodaning eng kattaqiymati bilan x -5 ning eng kichik
X i 11
qiymati ko’paytmasini toping.
32. x * У - — va 36 < — < 84 bo’Isa, * ning butun qiymatlari
ko’paytmasnini toping.
1. Tenglamani yeching.
.\44S,.V.V.VA\S,A4V.V.VA.
Л'.’.'.'Л'Л'Л'ЛЛ^.уЛУ/.УЛ’Л'Л
% *Л,.,.-Л,Л*Л,.у.*Л'.'Л Л Л,А\Ул
V.V.V.VAV.V.'AV/.V.V.V.'.’A*. .
• 4\S44W.V.S%VAV.W.VAW.'. «
w
2) Vx^-Тз - V5j^3 = 3 ;
<\SVCW.VeV.V. .
1 •
ViSVftA
6) -j3x + 2 = 0 ■ 7) = 3 ;
10) -У2ДГ-1 =3 ; 1 LjjiVJ+T = ЯВ
1) 4x-b = 2 ;
5) V2JT-1 =0 ;
9) Vl-2* = 4 ;
12) v .v - 2 = V3a' - 6; 13) V*2 +24 = Vl 1дг; 14) л/л" + 4л = л/14 - л;
15)л/л+2 = л; 161 л/Зл + 4 9^Тт1В80-л2 = 2л :
- .• VA^VA1*. Wi ’vlvlviv/lv!'.
18) V0.4-X2 = 3л; 19) 20) V*
'vXvXv
УЛ ’•SAV.v.r.y.v.v. •\V*y*y
+ X-6 = X-1
21)
23)
yj x2 — 4x + 9 — 2.x - Slit's ,22) ylх%ШЗх + 6 — 3.x + 8
2x = 1 + V77i ;; 24}::::%l|l||/l3 — 4.x = 4 ; 25) yJx + \2 =2 + J~x ;
л/4 + .х +
WAW.V.SWA
/I '1Щ1.
= 4
wi ..
w;
IWb
yJlM+1 + •yj'ix + 4 — 3 *
26)
28) V4*-3 4f3S£:#4^y29) л/1^7 - л/^+17 =-4 ;
. .v.vav. . v/!vXv/Iv/Xv. !v!vlv
V.V.VAW.V.Vi
30) 31) л/З^ = 2 ; 32) V3
д: +1 = 7
тШЖ; 34) V5JC-1+3X2 = 3* ; 35) л/2^Т = * - 2 ;
3 ; 37) J2x
+ 5*-3 = .x-f 1
38) л/З-Х2 — 4л:Ч-2 -х + А ; 39) V* + l 1 =1 + л[х ;
40) Vif l9 = 1 + л/Зс ; 41) л/* + 3 + л/2х-3 = 6 ;
42) л/7-Гх + л/Зд:-5 =4 43) л/Зх-5 + 7 = 9 ; 44) 10-л/бх-5=5
45) л/з+Ж^5=2; 45) 1 -л/х+Т = л/2х + 3 ;
46) 8-у[х + 2 = у/Зх + 4 ; 47) 2 + y/lO-у = <j22-y ;
SOLIYEVX.X.
#28#
01/11/2015 у
***** V KITOB *****
48) л/з^Гу = V5-; 49) у1у=у1у2-\; 50) V*2 -13-л/З = 0 ;
51) х-2 = у/2х + 7 +2 ; 52) y[y + 4^[y^\-6 = 0 ;
53) V6*+9-1,5^11-*= 0 ; 54) ^Ъу-2+2 = 2^+2 ■
55) ^1 + Уу[у~ +24 = 1 + jv ;
56) З-УЗ* + 5 = 2у[3х2 + 24
57) yfy + i = 2j3y-5-2 ; 58) ^~Уу[у=У
Ж
59) -Jl + 3y= -j== ■ 60) V^ + 5 + л/2^+8 = 7
61) л/Зс^З- - 73дг - 2 = дг2 + 4 ; 62) /5^+7 - л/зЗГГТ = Jx + f9
*,vl
63) 2 + yjy + 5 + yfy = 0 ; 64) л]2- у - 7 + л/Jr 5 = 0 ;
'VN,.
.лк %
65) Vjc-7+3 = 0 ;
68) Зх -1 1>/х + 6
AjXvKS
v.w.v.
■Six'
' * - 7
' ; 66) Vl2 + V2^-7 = 11|Д]|)| ,f»l = 0 ;
' S\;.VAy.yAWAW.V.VAW.SS . ..
vkv*v.v.v.*.V.VA\%v.eA%v.%vw.
= 0 ; 69) 3J + 6 = 1 oVj+T ; 70| JP4-4 = 5^/ -2
'S
+ii
73)T'2 -4y + 10 = 3/y2 -4X41||3« P^h
,»Ж ШШРЧ ' I*:, Л/ V
,# - 1 #'
_ Xviv’-v’-y *:•:># x* W
2 . -fiiil6 l|ip2. a.
, -5 ХЖ "гШмШь, 76) „
1
J у-з]у2+у у+з]у2+у ’
jjk 78)-y + 2 + ^4 + 2y-y
79) J—Й!4B||2It5^ j 80) У+ 2-2^5 + у = 0 ;
81 ы8Ш
SM
7
о
= 7.
85) /у--Т + 6i[y^\-16 = 0 ;
86)
_ 3 + \[y = l]\ 2(y -1) ; 87)
2 + 2j I 2 +у
2 + у у 2 + 2_y
J7_
12
88) л/Зл: + 4 + л/х-4 = 2л[х ; 89) л]х + \[х + \Л + ylx-yfx + U = 4
90) y/l5-x + л/3 — х - 6 ;
91)
1 + Jl + xylx2 -24 =
92) yj3x + 7 — Ух+Т — 2 ■ 93) yfl^^x + \Jl--^Jx — 2 •
SOLIYEV X. X-
#29#
Ol/H/2015 у
***** у KITOB *****
94) 2V7^x : 0,бД = 10VT5 :^л/216л/9 .
95)
x + 5
\
/
+ 4
V x
\
\~
\
* + 5
= 4 .
96)
\j 24 + ~ + л/я = 1
97) V* + 34 - Mx-3 = 1 ; 98) *2 + 3*-18 + 4л/*2 + 3*-6 = О
99)
л/*2 +32-2V*2 +32 =3 ;
'\fx-4yf.
101) .x2 + yjx2 -ь 20 — 22 ■
103) л/-£3 +8 + ylx* +8 =6 •
105) ylx2 +9-ylx2-1 =2 ;
104) л/лг + l -JA-X = у/2х-\2 ;
106) л/ЙМ
1№Ш№тт
2
'л
V.
'.,a%s,a\T,X'a\vX,X\vX\v.*. ...
S’.V.V.VA'.-.'.N-.V.W.WV,
л/з
+1 + л[х
2 + з — V бх2+10
.-ЙЙ
AW.V.V
..........
107)
108) л/^ + 2 + V3x + 8 = yflx + 6
109) л/2лг + 5 + л/5лг 4- 6 = Vl^^+^5
ПО) *2 -4х-6 = Vlx2
йШ:
"а
%
dr. 3
.. AW'.V.S . . .
y^w.v.w.ss
•.*Л*Л*Л\Ч*.\Л*Л,Л'Л
‘V.S'.V.VAWA’AWAW.V
* •.S'.SV.V/.V.V.V.S'.S’.V
• V.4*.S‘.VA*.\\\%VA'
v.
P
lv-
ЩШ
5
«=■
wmmm*.
'-'ШЖШ
.•.v.v.v.*.
ш.
AV/i’i’.V. WI'X
у/, л1 •
.. л*л*.
em
m
хЖх
ШуУ
112) V5
VA'A'.VAW.
WA'AWA’A’, ч,
W
•S4V
.X-S:-.
AVA'.V
W.V.VANVA
XvX£X;XvX£.e
‘yiyAX^Viy/A
xx-x*x::x>x
lillk.
2. Tenglamani yeching va quyidagi shartlarni bajaring.
л^л**;л'л\у
i уig’indisini toping, ylx4 +5x2 = -3x
^(x + 2) = .x + 2
' .VAWA'.V.VA'.
•.•.•.y.V.V.'.'.V.S’.VAVVsw.VA*.
N’AViNV.SyS’.V • WAV.
• • 4\VV>.4S ' w.y.
VWNVS
A’A*.
sistemasini yeching.
%
ЙУ
-WvX'
чй:
3) Tenglamalar sistemasini yeching.
4) Tenglamalar sistemasini yeching.
Жх~2 У
^(x + 5)2 = л: -h 5
yl(x-5f =5-x
yj(x + l У = x + 7
yl(x-7f = 7-л:
SOLIYEVX.X.
#30#
01/1V2015 у
***** у KITOB *****
л](х + \)2 = x + l
5) Tenglamalar sistemasini yeching. 4 ^ _ x _ j ;
V2 2
x -Ъх + 5 + х =Ъх+1\
M
mm
2$*.
7) Agar V*4 -9x2 = -4x tenglamaning katta ildizi *0 bo’lsa, *0+Ю
nechaga teng?
8) 4a -4b = 4 vatf-6 = 24 bo’lsa, 4a + 4b nimaga
9) 4a -4b = 1 va a -b = 35 bo’lsa, 4a + 4b nimaga
10) 4a + 4b - 6 va a-b = 48 bo’lsa, 4a-4b nimaga teng? %
11) 4a + 4b = 11 va a-b-176 bo’lsa, л/л -
12) 4a -4b = 9 va a-b = 63 bo’lsa, 4a 4
13) 4a -4b = 8 va a-b = 96 bo’lsa, + 4b
14) 4a -4b =15 va a-b = 15 bo
15) 4a + 4b = 13 va a-b = \82
mm,
v.e.w.4\y
WAV.V.'A
'.’.‘.•.■.V/.V.
V.V.V.4V
Л%
nimaga teng?
AVVAVVA
ASV.y.V.SNVV.'V.V.
шрм :i
. A\4\WA\4VSW.V.<4
5 1
SWiVAW.V.vA
W.V.V.'.V.'.V
* v v»v»v
+ V* nimaga teng?
-л/б nimaga teng?
• . m. v 2
16) Tenglamaning natural ildizlari nechta? v(3* -13)' = 13 - 3x
•Xvlvlv 'vXvIvXvivl'X;. Xvlv.'v
17) Tenglamaning natural ildizlari nechta? л1{5х-21)2 =21 -5x
.V.WA\SSNW.¥A\S,.W •
- • VA\\\4\VAVAVWAW.\
* Ч*Л*.\'Л,Л,А,А,А-Л*Л'.
■ ■ •> .л .x.v«v« • • ■ • «
AV. * SNWA'AWA’A’AWA ■» f
^v.v. * I у v л
18) Tenglamaninginflural ildizlari nechta? yj{2x-39)" = 39 - 2x
.WAV
Vi*
20) Tenglamamrig natural ildizlari nechta? л]{\2х-207)' = 207-12л:
>*л; %¥:*>5й’й-й:й-
Л 4 f l'< X|Xv ^ 'vX, xy^>x^>#x>y ^ ^ ^ • _ • _ • ^ _ •
21) Tenglamaning ildizlarining o’rta arifmetigini toping.
>Xy ‘A'.y^’A’V.y.y.; VA
22) Tenglamaning ildizlarining o’rta arifmetigini toping.
• |jjc + 2) • (* -10) • 4s-~x = 0
ko’paytmasini toping.
*- +77-3 = 0
24) Tenglama ildizlarining ko’paytmasini toping.
V*2 + 220 - 3JV*2 + 220 - 4 = 0
25) Tenglama ildizlarining ko’paytmasini toping.
4x2 + 609 - V*2 + 609 - 20 = 0
SOLIYEV X. X.
#31#
01/11/2015 у
***** у KITOB *****
26) Tenglama ildizlarining ko’paytmasini toping.
V*2 -15 + 5V*2 -15-6 = 0
27) Agar a/25 + -/* + 13 - 2 = 0 bo’Isa, + ~ ning qiymatini toping.
27) Agar \/23 +V^+W-2 = 0 bo’lsa, ^ + ^ ning qiymatini toping.
28) Agar л/238 + -\/д + 16 -3 = 0 bo’lsa, >/* + —
29) Tenglamaning ildizlari yig’indisi 10 dan qancha kaml
V*3 -2jc2 -4.x = .x
ning
toping.
ччччччччч
х*хх:х*:
30) Agar
1 1 * И
1 6 bo’lsa, 6- + *
XX о
AW.
Ж.
*$Ч;
•ччччччччч
ччччччччч
.•y.V.y.VAV.'.y.V.V.VA
31) Tenglama ildizlarining yig’indisini toping. л/д + \W\l2x + 3 = 1
32) Tenglama ildizlarining yig’indisini toping. (l6- x2)j3-x = 0
-Чч'Хчччччччч'Хчччччч у v
/ ~ \ I
33) Tenglama ildizlarining yig’indisini toping. (■* - 9)Jx + 2 = 0
“ :x-:4x-x •
. (x2 - Зб)л/д + 5 = 0
35) Tenglama ildizlaripj|g?yjg’indisihi toping, (д2 - 25\jx-4 = 0
ччччччччччччччччччччччччч _
• * \yAW.V.4*A*AWA,.V.^
* ЧЧЧ\ЧЧЧЧЧЧЧЧЧ ЧЧ\Ч*Л'.Ч*.
36) Tenglamaning natural ilclizlari nechta? у(2д-7)4 =7-2x
чу.чччч*лчч:л*
WWeHW.N'
ЧЧЧЧЧЧЧЧЧ’
ЧЧЧЧЧЧЧ*
37) Tenglamaning natiillldiztari nechta? V(3д-35)4 = 35 - Зд
\ЧЧЧЧЧЧЧЧЧЧ\ЧЧ
444444444444V
38) Tenglamaningmaturai ildizlari nechta? v(5*“ 6^)4 = 62 - 5д
39) Tenglamaning ildizlarining o’rta arifmetigini toping.
ЧЧЧ44Ч4Ч4Ч4ЧЧЧ Ч4Ч*
чччччччччччччч 4*
v.y.v.-,
Ч444Ч4Ч4ЧЧ4'
W- 4 = 0
WA'AWAVi'A * *.V.*
ЧЧЧ*
ччччччччу.ччу.у _ w
Tenglamaning ildizlarining o’rta arifmetigini toping.
д - 6л/д + 8 = 0
41) Tbhgiamaning ildizlarining o’rta arifmetigini toping.
д-7л/д + 10 = 0
42) Tenglamaning ildizlarining o’rta arifmetigini toping.
д-7-Уд + 12 = 0
43) Tenglamaning ildizlarining o’rta arifmetigini toping.
x-9y[x + 8 = 0
44) Tenglamaning ildizlarining o’rta arifmetigini toping.
SOLIYEVX.X.
#32#
01/11/2015 у
***** V KITOB *****
х-&л[х + 12 = 0
45) Tenglama ildizlarining yig’indisini toping. V-*- V*4 -13*3 +36 = 0
46) Tenglama ildizlarining yig’indisini toping.
(x- 2)‘yjx2 - x-20 + \2-6x = 0
47) Tenglama ildizlarining yig’indisini toping. 6 + V*2 -3* + 6 = 2x
48) Tenglama ildizlarining yig’indisini toping.
x2 + 5* + yjx2 +5jc-5 = 17
49) Tenglama ildizlarining yig’indisini toping.
x + 4j^\ + 5 + J\& + 6j9^-x = 9
50) Tenglama ildizlarining yig’indisini toping, (* - 25)- V6 - 2.v - 0
Xv!\ a ’X*!
Jf
л/?+4-7| = 2
Vjc-6-15 =11
Vjc-7-9=6
v-v’ . .
56) Tenglama i 1 diz;i^||ning^^p|mdisini toping. (4- x2)V“l-3x = 0
луДу*У.
x + \Jx + ... = 4
WAWAWvVAW< v+v.v.v.v.v
• Чй®Ш§й§&; Шш? I ,
Ч&Д Г-Шу -i / -л /
шшуттФ+У^^
* + ... =5
v.v.v.v.v.'.-.v.*
. V
.‘X'.V.'AV. v.v.v.v.v.v.v. v.v.v/.v
.nsv.v.w.va .w.vAmyAV v.y.y.y
K'.W.SW.SV, V.4%y.y.y.y. .y. I
Л * 4WA чу/лулу.у.ул, V.y.y )
V
X +*<J.
x + \jx +... =3
* + л/
x + \lx + ... = 4
tJx + \[x + %J.
x + ... =2
sss
]Jx + \fx~+\lx +... = 6
д/jc + y[x~+ y/x + ... = К
-/* + -/* + *Jx + ... = 8
^Х + л[х+у1х + ... = 7
-/Ж# ^
#33#
Ol/H/2015 у
***** у KITOB *****
66) Tenglama yeching.
yjx+y[x~+^
x +... = 13
67) Tenglama yeching. yjx-ylx-l^Z = 8
68) Tenglama yeching. V* • \jx-\f~. = 3
69) Tenglama yeching. \jx-\jx-\f~ = 5
70) Tenglama yeching.
\jx-yfx~-
= 6
71) Tenglama yeching. Jx-Jx-JZ = 15
72) Tenglama yeching.
73) Tenglama yeching.
74) Tenglama yeching.
yjx->{x~-
= 12
= 24
S 27
«ж-
I ■■■ ■ ———
/ /
75) Tenglama yeching. V* * v* ■
dll
• wy
78) Tenglama yeching. У * ЩЩШЛ1у 2л/2
Лчч
79) Tenglama yeching = 3v3
лчулчу
У^ШЖУ • = 5V5
V.%*.4%V*.VAy,
**.v.v. v.v.v.O’.s
лу.улу.УА'Л
'4vivA\v!s,A*A
\V»NV.4V.V.V,'.
♦
.Л’Л’А’Л-, ^
ОСЛЧЧ’АЧЧУкЧЧ. W.V.V.V.V.V.V.
ШУ
•%y
.'л*л*.*луд
‘.УДУДЧ
V.V.'/.W.'.'.'.SS»
WAV.’.S'.V.V.
J'-V.v
•Vw^
= 3
= 7
’.'.•.VA'.y.VA
VWA’.SW.V.V
V.V.'.V.'.V.VA
•AV,V,AyV.V.'
y.£.£.y.y.y.
v
WA'.V.y.V.SVNVV.
жтщт&. 4 *- - v
у7уу!у7у- =1°
= 49
^*2-V*2-V?I=81
42-V^-V?I=16
85) Tenglama yeching.
86) Tenglama yeching.
87) Tenglama yeching.
88) Agar л/8-л + ^5 + a = 5 bo’lsa, ^(&-a)(5 + a) ning qiymatini
toping.
SOLIYEVX.X.
#34#
01/1V2015 у
***** V KITOB *****
89) Agar л/З-а + л/б + я = 7 bo’lsa, yl(3-a)(6 + a) ning qiymatini
toping.
90) Agar л/9-х + л/S + x = 9 bo’lsa, yl(9-x){S + x) ning qiymatini
toping.
91) Agar Vll-* + Vi8и- д: = 11 bo’lsa, yl{l\-x\\S +x) ning
qiymatini toping.
92) Agar -f\A-~x + л/23+"? =17 bo’lsa, yl(\4-x)(23 + x) nitig qiymatini
toping.
93) Agar J25-X1 + JlS-x* =5 bo’lsa, /25-x2 -/l 5 - *"|l|j|l
qiymatini toping.
94) Agar Vl3 + z3 + Vz3 -14 =3 bo’lsa, Vl3jj3|-Vz3 -1| ifodaning
qiymatini toping.
.VA\y.SS|.
A’.V.V.VAV.V.'.'.’.SW.'.NV.V.'.-. _
• • • sw.v.*.w.‘.,.s,.',v.,.v,v.y.y.v.4.
i ■ ■ > 111 м и >i
• N^y.S'.S'.S'.S'.V.V.V.'.WAWV
llli
V.V.WAWA*A\V.V.V.V.y/.,A
-V.SVV.V
95) Agar л/l8 — л:2 + Vl4 — д:2 =4 bo’lsajVl8 — л:2 ifodaning
qiymatini toping.
96) Agar л/19 + я3 + y/a3 -29 = tilbo’lsa, лЛ.9 + я3 - V<r3 -29 ifodaning
qiymatini toping.
-AW-’
•г’:-:-:-:-:-:-:-:*:-:-:-:*:-:*:*:*:*:*.
97) Tenglama yeching. ;
98) Tenglama yechingfllfl
.йу!;л;л\
•.•.V.VAWW.SNS V.VA*A
'ЩШШ. #
£& 2#§t
•N:*
лч
<W.S4y.S»vX
v.v.s'.v.v.v.v.
Ч’.'ЛУЛ’АГЛ'ЛЧ
VAW.v№.4*A\
.S'.V.V.V.
AW.V.VA’,
y/KSWAV
(A'.S'.V.V.V.S*. ¥
l.*.S*.e.\V*X\S\\V« ...
V.V.V.VAVXS'.'AI
хщщ^х=5
y[x - 6л/? + 8 = 0
101) Tenglama yephing. liv? - 5Xfx + 6 = 0
'чтщ& щу I— ъл /— л л
.* у/х + 8Цх-9 = 0
NV.4V.V.4
V.VAV.W.
: • ‘ • •
4~Х -Зл/?-4 = 0
100
ning qiymatini toping.
25
105) Agar V? + 6V? = 7 bo’lsa, ning qiymatini toping.
810
106) Agar \[x-5\[x = 6 bo’lsa, ning qiymatini toping.
107) Agar \fab = 2л/з va я, beN bo’lsa, a-b qandayqiymatlami
qabul qila oladi.
SOLIYEV X. X.
#35#
Ol/H/2015 у
***** V KITOB *****
108) Agar \[ab = Зл/2 va a, b e N bo’Isa, a + b qanday qiymatlami
qabul qila oladi.
109) Agar yfab = 2*j2 va a, beN bo’lsa, a-b qanday qiymatlami
qabul qila oladi.
109) Agar ifab = 2^5 va a, beN bo’lsa, a + b qanday qiymatlami
qabul qila oladi.
110) Agar Xfab=Ay[2 va я, beN bo’lsa, a-b
qabul qila oladi.
111) Tenglamaning nechta ildizi bor?
mm
ЩЖ
x-xvix-x:::::-:-:-;
v<v*V.V.V.V.V.
112) Tenglamaning nechta ildizi bor? yj(x - 4)2 + \/(9-*)3 = 1
113) Tenglamaning nechta ildizi bor? л/(2х^ЙйЕ+ д/(б-5л:)3 = 5
114) Tenglamaning nechta ildizi bor? +1
fc 4л:)3 = 9
115) Tenglamaning nechta ildizi bor? Ш(x- 4)2 + y(5 - 3*)3 = 7
116) Tenglamaning nechta ildizi t^r? vlll!§)2 + V(2* “ 9)3 = 15
117) Agar
Jx + y[y = 3
ЯШй-.
’em -tew.
‘ШШт. "*
n „ ХЩ1|
» %
Щ
1
\4x + Jy ~
18) Agar A|
• y.j.^.;.y.yy.y^WA\
bo Isa, * + у m toping.
1
•у
’Isa, * + y ni toping.
y:>.**: .;W
Щ|р9
— = 15 bo’lsa, * + y ni toping.
\[x~y[y =11
^ = 24 bo’ Isa, * + У m toping.
.yfx -yfy = 13
122) Agar 1 ^ _ 33 bo’Isa, * + у щ toping.
+ л/у = 7
123) Agar) ^ = 16 bo’lsa, * + y m toping.
SOLI YE У X. X.
#36#
01/1V2015 у
***** у KITOB *****
[ x-y = 21
124) Agar = 3 bo’Isa, x + y ning qiymatini toping.
I x - у = 45
125) Agar = з bo’lsa, x + y ning qiymatini toping.
f x-у = 63
126) Agar wj_j^_7 bo’lsa, * + y ning qiymatini toping.
x — у = 48
ж
#ii
127) Agar j г г- _ , bo’lsa, x + y ning qiymatini toping.>
lv* yy-o
тщттттм+.
.<ж>.
Г x- у = 99
128) Agar jyj_^ = 9 bo’lsa, x + y ning qiymatini toping.
Vx
х-х-й-й*:-:-
129) Agar
x-y = 165
= 4 bo’lsa’ x+y
vm
w
W
%
130) Agar
дг-у = 35
.v. v!\
Ж
4ft.
Is®
_ JZ = 5 bo Isa, if«nigiy
v ^ ... щфшш
•.•A'.v.w.swa
>>.
mng qiymatini toping.
/vX%vXv>>.
wv.yy
a® XW-rrfXvXvv, XV VA.
аммнвВнв
i bor?
132) Agar
v
W
. %р:ЙШ:хда::л
, * + */ s*iik,.
кЛь14Т“-
?1
у = y\6-x
y-x = 4
-Ji
x + y
ning qiymatini toping.
133) Tenglam^ing haqiqiy;i 1 dizlari ko’paytmasini toping.
V2
л:2 +
VAfAWWA'.'.V
\*A‘A\V.V.V.y.
- XSisIfk
Г. Л | AV.V.W,-
v.\\>y
»w/. XWX'AVA'X'Iv. 'V*AvSAv‘*\4V •
134) Tenglamaning ildizlari ko’paytmasini toping.
Z+yy--'+C+
x +
111Ш12
:.+
. AV.VA'A’.VAA VAVAV.VA'AV.V.
-5 = 17
ildizlari ko’paytmasini toping.
V*4,10MJl + x* +^2 + x2 -2jx2 + 1=4
136) Tenglama nechta ildizga ega?
x-9
Jx +3
= x-\5
137) Tenglamaning nechta ildizi bor? *Jx-5 +3jx + 3 = 10
138) Tenglamaning nechta ildizi bor?
“\Jx + \ — yJx + 7 + “\J 8 + 2 V x + 7 + x — 4
139) Tenglama nechta haqiqiy ildizga ega? V*2 +1 -ylx2 -1 = 1
SOLIYEVX.
#3 7#
01/11/2015 у
***** у KITOB *****
140) Tenglamaning ildizlar sonini toping. л/2-х2 -Vx2 -4=0
141) Tenglamaning turli ildizlar sonini aniqlang. Vл/1 l*2 +1 - 2x = l-x
142) Agar Vx - 3 - Vx +1 +2 = 0 bo’Isa, x3 - 2л: +1 ifodaning qiymatini
toping.
143) Agar л/Зх2 -6x + 16 = 2x-l bo’Isa, x2 *(x + 2) ning qiymatini toping.
144) Agar Vx + 1 + x -11 = 0 bo’lsa, x2 -11 ning qiymatini toping.
145) Agar у1х + 3-у1х + \4+у1х + 3 + у/х + \4 = 4 bo’lsa, —7 ningftymatin:
toping.
■<шш
146) Agar yjx + 3-Jx +14 + V* + 3 + Vx + 14 g||4 bo’lsa, x(x+l) 1 ifodaning
qiymatini toping. jgfcL. Ц
I . I ,
147) Agar vl + Vx -1 + ф - Vx -1 = 2 bo’lsa, ning qiymatini
tnnina
toping.
л[бх
.•Л'Л'Л'ЛЛЧЧу. \vy.;
Ч*Л'Л* Л*Л‘.‘Л .•••.V.
•AV.W.VK\4WSV .W.V •
N'.'V.NVANV.NW.V. Л*Л*Л
149) Agar a3 + 5V<r3 +1 -13 = 0 bo> IsaJlVtf3 +13 ning qiymatini toping.
150) Agar x - Vx + 3 -17 = 0; bo’lsa,;- Vx + 3 ning qiymatini toping.
151) Ushbu ТзТзЩв-х tifgk/x ning qanday qiymatlarida o’rinli?
152) Tenglik |§ning qanday qiymatlarida to’g’ri bo’ladi?
.wX'Xw. \v.v!v'.v!v!v.*.
.■A4SW.\SV.SV .•.W.V.
ЛЧЧ'Л'Л'Л'Л'Л’. W.WA*. .V
‘А’Л'ЛЧуЛ’Л’Л*. WAV.V
AWAWAW-y.. ' <*.44
ЧЛ4444,Л.4'ЛЛ. 4*
*.4WA44,.44444
4у.4'Л4уЛ444‘.л,л..
2x -1)2 (З - х|Ц:;;(2х %
to’g’ri bo’ladi?
Jir 0-
‘ V«V.V V-VA'A'.VAWA V.V.*.*
N’.'.’A'AWAWA
4%444,A4V.V.y.4
44444%4W.444,A
•.•.•.•ЛЧЧ’Л'Л'Л*.
llllilMlp^2{a-З) va ау/3-а ifodalar qaysi oraliqda aynan teng bo’ladi
л w‘‘у-:’:* A A * A • • * * A • • . •
155) Tenglama ildizlarining kvadratini toping.
VTV? ЩТ+гЛ^х +i = o
л/2 - x
156) Tenglama ildizlarining o’rta arifmetigini toping.
2-x
3 + x
157) Tenglamaning ildizi 12 dan qancha kam? л/б + х - V8-x = 0
158) Tenglamaning ildizi 15 dan qancha kam? V5 + x - Vll-x = 0
159) Tenglamaning ildizi 23 dan qancha kam? Vl3 + x -V39-x = 0
SOLI YE V X. X.
#38#
01/11/2015 у
***** у KITOB *****
160) Tenglamaning ildizi 9 dan qancha kam? Vl8 + * - V32-x = 0
161) Tenglamaning katta ildizining eng kichik ildiziga nisbatini toping.
V*3 +19 = * + 1
162) Tenglamani yeching. у/Зх-7 - ^7 -3x = 0
163) Tenglamani yeching.
164) Tenglamani yeching.
л/l32 -122 =^625
Ъу[2х - 5л/8х + 1у[Пх = 28
165) Tenglamani yeching. V*+2V*-T - л]х-2^[х~^\ =:£ к
2y[x - y[2x
+ 3 — Гх +1
sllllllil::::::,,,.
• 4%%\W%4\\VW,Ay.,,N,A.V.V.'.
ill
y/x2 -x-2 = x-3 Ш
Cv2 - У Уд- +1 4lllfe.,
'*л*Ачу.'.;л,л,л,л,,ч,л,л,л
166) Tenglamani yeching.
167) Tenglamani yeching.
168) Tenglamani yeching.
V— —1 /
x2 -Ax +A = yjx2 — 1 Олт ч- 25
170) Tenglamani yeching. y/5-AMk5XAx
171) Tenglamani yeching.
172) Tenglamani yeching.
173) Tenglamani yeching.
•\\vlv.*. .-.v.
.WAW.V.V.*
V»
■ • • • • ^ •
»\v.sv
.\v.. .
s&m
л: = 4
ШЩШ
174) Tenglamani уееЙп§ l!!!|/^ +1 + y/2x + 3 = 1
,2 + 9x + 5 -^2x2 +л:-1 = y/x2 -1
*Jx2 — 4л: — 21 + л/l 0 -H 3x-x2 = 2
WiWAW
\^\<\WAvX\
*
V.V.VAW.V.V,
•лулулу.ул*.-
'•.y.V.yA*№AV.
V.S’.V.V.VV.'.V.-
SNSWXWAW.
»;л;.у.\у
m. ®
v.vw. w
w
•.•лчулу/л\«.у
-2,5 = 3
x-\
‘J7-*J7+4J7-\I7=7
/л о /? , /T c /T
179) Tetiglamani yeching. yjx - ylx + ylx - \Jx = 24
180) Tenglamani yeching. V? + V*3" + \l~x* + \[x* = 15
181) Tenglamani yeching. V? + V? + y[x* +yfx~ = 34
182) Tenglamani yeching. yfx2 - yfx* + V? - \lx* = 42
3. Tengsizlikni yeching.
1) 4x-2>3\ 2)V3--4oc<6; 3) ylx2 +3* < 2 ; 4) ylx2 — 8jc > 3 ;
SOLIYE V X. X.
#39#
01/11/2015 у
***** V KITOB *****
5) л/х-2 <1 ; 6) л/ 2-х >x; 7) yf2-x<x ; 8) л/5х+ТТ>х + 3 ;
9) л/х + 3 <x + l ; 10) л/х2 +3x < л/2х + 2 ; 11) л/Зх^2>х-2 ; 12)
л/2х + 1 < x-1 ; 13) л/4х-5 < л: ; 14) л/х + 2 > x ;
15) yfx*~+3x <x-2 ; 16) л/2х + 5 <11; 17) л/2х + 4 <x + 3 ;
18) л/х2 + x-2 > * ; 19) л/х2 -4x > x-4 ; 20) л/х + 78 < x + 6 ;
21)
23)
л/х2 -x-12 < 7-x ; 22) л/х^-5~x + ~6 < 2x - 3 ;
V*2 +4*+ 4 < x + 6 ; 24) л/х2 + 2x + l +
% "W
«ft -v ••
.w.v.v.
W
25)5л/х<х + 6; 26) лЛбх2 -24x + 9 < л/4x2 + 12x + 9 ;
27) Vx+T < 4 ; 28) л/Зх-8 > л/5-х ; 29) л/х2* + л/х1 < 4 ;|f
30) V7 + 2V7<9; 31) V7 + V7<10
\ v^y\\\\44v.^s\\w.y.s\\v,
^eir
<18
33) д/|х-3| + 1 > 2|x-3|-1 ; 34) V3x +10 > л/б-Л||Г
2*Л\\*
4. Tengsizliklarni mulohazalar yuritib yeching.
•w-v.'.-.
уЖ;::
;i|s.
1) 7771>-2; 2) |* £7*2-2*-3 <0;
4) Vx-2 + Vl-x >3
.'.VAV.V,
1 д
r-fl5>0; 6) v+v - лТГ-7 > 0
7) (* -1)77 - л -Ца; "IJ (з#*)7
x2 + x-2 < 0
9)
X-7:
vK .* SBSK>n
УлУЙтайЯл
vwCv^.v.
1 u
V4x2 тшШх Я1
чйгйй
7
< а
|х|
vlv
л/2
х2 +15х-17
IwXv^
XvX-XvXv>№. •
V.V.V.V.V.V.V
V.V.VAV.V.y
л
л*м
о
12)
10-х
х2 - Зх - 4
л/х2- 4х + 3
>0
<0
\;.,.4,.ssvoAS\v/.WA;.y.y.;.%y.\ * v.v.T.v.%v.\y
шшшррйл
л/6 ч
|Ир^Щ
HWyANN»1 ;лу.^
+ X — X
< 0 ; 14) (х + 3)• л/х2 -х-2 > 0 ;
15) (х- 3)• л/х2+х-2 < 0 ; 16) (х-2)*л/з + 2х-х2 > 0 ;
17) (х + 3)-л/Ю-Зх-х2 >0 ; 18) л/Зх- 8 < -2 ;
19) л/5х-2х2-42>3 ; 20)
(х -1)2 (х - 2)
л/х + 2
>0
21)
л/б
+ X — X
>
Тб
2лг + 5
+ Х-Х
х + 4
22)
2х-3
5х + 7
> -2
SOLIYEVX.X.
#40#
01/1V2015 у
***** V KITOB *****
5. Tengsizlikni yeching va quyidagi shartlarni bajaring.
1) к raqamning qanday qiymatlarida V30 + k ning butun qismi
5 bo’ladi?
2) n raqamning qanday qiymatlarida V49 + n ning butun qismi
7 bo’ladi?
3) к raqamning qanday qiymatlarida V73 + к ning butun qismi
8 bo’ladi?
.-xv::x-.. '<•>
4) n raqamning qanday qiymatlarida л/173 + л ning butun qismi
14 bo’ladi?
5) Tengsizlikni qanoatlantiruvchi butun sonlar nechta?
Ill
1) -Jx + 2 >x ; 2) л/5-х2 >x-\ ; 3) Jx2 -6x + 9 <3
л[х +
Зл: - 4
8-*
>1
;5)
2 • (л -1)
(* + l)J
< О
6)
w2
:!Г X
midrn
x-2
<0-
7) V^-50-Vl00-Jt >0 ; 8) 2 ; 9)Vjc-4-V*-7 >1 ;
5-V* ^ л
I0)XT 0
-2
<1
13) *J& + 2x-x2 >6-4
6) Tengsizlikni qanbatlantiruydhibutun sonlar yig’indisini toping.
1) yfx > x-6 ^
5)x
ix2 +
.же,
6)(.
л*А\\*А‘л
I.*
'.Ч’Л'Л'ЛЧЧЧ'.Ч'Лу*
ч" УС-
‘•’•Л** *•*•'•*•*•
3) Зл/Gc > * - 4 ; 4)Jx>x-12 ;
улхл
tlr > о
JC +
*2 >0
V.V.VA'.V.V.V
•Л'Л'Л'Л'ЛЧЧ’Л
P/.W'.’t-.V.VAV
-6.t + '))-V36-.v: >0 ;
6л: + 64)-л/81-л2 > 0 ;
У-^Jx + 5
10)
9)
х~ - Aj'^JT—x ’ V дг — 18
7) Tengsizlikning butun sondan iborat eng kichik va eng katta
yechimlari ayirmasini toping.
1) x-4sfx - 5 < 0 ; 2) x + 6y[x-7 < 0 ; 3) x-3-Jx-4 < 0.
8) Tengsizlikning butun sondan iborat eng katta va eng kichik
yechimlari ayirmasini toping, y/x2 -16 < V4* + 16
9) Tengsizlikni qanoatlantiruvchi eng kichik natural sonni toping.
<0
8-jc
>-l
П)
y/2x + 7
6-3*
>0
X.
#41#
01/11/2015 у
***** у KITOB *****
1) V*2-3* + 2 >0 ; 2) y[x^-Vx--8 > 0 ; 3) V?+4x-5 > 0 .
10) Tengsizlikning eng kichik butun yechimini toping.
2-3*
* + 4
> -2
I) -j\2-x < 2 ; 2) Vl7 -дг <5 ; 3) J
II) Tengsizlikning eng kichik musbat butun yechimini toping.
л/* + 5 л
\-x
12) Tengsizlikning butun sonlardan iborat
kichigining yig’indisini toping.
vaeng
-.•.•.v.v.v/
•.v.v.y.y
• V.V.V.'.V.VAV/.V.V
%
N’.V/.V.
13) Tengsizlikning eng kichik butun musbat Aeng katta butun manfiy
yechimlari ayirmasini toping.
••2 -2*-8 л x2 - 5x - 6 л;ё
2)
1)
>0
* -*:-'
.WAW.V.*.
,<VAW\<4yX\V.*. VA
■ASS'AVAVAV.'.VAVV. V-'
vv<\
A’A'A’AVaX'A.VA... •
SW.S'ASSW A'A'.y.V.
V*2 f 9
>0
yfx2 +1 ’ V^+4
14) Ushbu yJ9-x <2 tengsizlikning yechimlari OX o’qidajoylash-
tirilsa qanday uzunlikdagi kesma hosil bo’ladi?
15)
1) /(*) = yll-2y[x^ ; 2) f(x) = л/9 -2ylx + 2 ;
VA'AV.W.y.VA’Ay
3) /W-V 12-зЛ|^ Tp[*)=V28-4V^5.
16) л ning qanda> qiymatlarida .v = v'2.v - I funksiyatiing qiymatlari
3 dan katta bo’fm^iifl^1
. XyivivAvi’A’A
л% Xv!\v/a;!v!vav. wavwv.;>.\^ .vn\v
17) * rling qanday qiymatlarida у = л/З* -11 funksiyaning qiymatlari
•.\w.v •
IT
SOLIYEV X. X.
#42#
Ol/H/2015 у
***** V KITOB *****
SONLI KliTMA-Km UKLAll
zi
1. Sonli kctina-kctliklar.
1. n - hadining formulasi bilan berilgan ushbu ketma-ketlikning dastlabki
5 ta hadini hisoblang.
l)tf„ =2/7 + 3; 2)an=2 + 5/7; 3)a„=56-3/72 ;
An -1 1 , ^
= n2 -2/7 + 5
4)
5)
=
6)
5 ’ " /7 + 2 ’
2 A • vv
2. Ketma-ketlik a„=n - 2/7 - 6 formula bilan berilgan.
1) -3 ; 2)2; 3)3; 4)9; 5)-6; 6) 4; 7) 18 soni slC1"
ketma-ketlikning hadi bo’la oladimi?
3. a, = 3 Sharti va 1) =3a„-l ; 2) ажЯА- 2a
*•*•*#*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*•*§*•*§*§ • **^»*в*•*•*•*•"
3) =«,;*3»-7 ; 4) = (a„ - 5)2 +in^Ppp=7 - 3/
rekurrent formula bilan berilgan ketma-ketlikning dastlabki ta hadini
. • аЖШ&х s;&- :й:‘‘
toping.
4. Sonli ketma-ketlik п - hadining я„ -:11|1УХя + 4) formula bilan
jOinwOT. ^ y-v..
3) a„ = 84 ; 4) «„=14;
ШШШ*.
»»Xv.
ШШШ.
.v.v.sv..
.•.•.S'.V.V.W.
%
berilgan. Agar 1) an = 150^
bo’Isa, n ni toping.
5. #„+1 = A rekurrent formula va;ii, = 256 shart bilan berilgan ketma-
pillk
ketlikning dastlabki 4 ta hadini toping.
Ш
.SV.NSSy*-
A'iSVSWiVV.
/
6. «, =1 sharti va 1) «Р*™
formula!
.v.v.v.
#388»
n
\
a
n
\
/
2) an+i = cos( ж/„) rekurrent
7.
iketma-ketlikning dastlabki 6 ta hadini yozing.
2
Я/1+2 ~an~ a„-\ rekurrent formula va ax = 2
al^lphiartlar bilan berilgan. Ketma-ketlikning beshinchi hadini
8. Ketma-ketlik n - hadining formulasi bilan berilgan. Shu ketma-
ketlikning (n +1), (/7 -1), (/7 + 3)s (w-2) , (/7 + 5) -hadlarini yozing.
1) an =-5/7 + 7 ; 2) ля =3(/7-10) ; 3) ля =2-3"+1 ;
/, \»+i
4)
= 7
1
3
/
5) a„ = (Зи + 5)(8 - In) ; 6) «„ = 5 -3”*2 + 1
9. Ketma-ketlik n - hadining formulasi bilan berilgan. Uning dastlabki
beshta hadini hisoblang.
SOLIYEV X. X.
#43#
Ol/H/2015 у
***** у к|tqв *****
1 )а„=п(п + 5)- 2) «„ = Зл_| -7(« + 1) ; 3) «„ = (2л - 1)(и - 8);
4) «„ = sm
6) «„ = 2"~2(3п - 4).
Л- _ яг
5) а» = 5cos— ;
« ’ 7 И ’
10. Ketma-ketlik «-hadining formulasi bilan berilgan. Uningo’ninchi
va o’ttizinchi hadlarini hisoblang.
2«-l « + 7 , ,
l)a» = 77j; 2) = 7—7; 3)а„=|4и-15| + 23 ;
Зи-2
4) an = 45 -14 + 5«| ; 5)
=
_ |7 - 2«|
4« +1
6) a„ =
* • • W,4V.,.‘.'.4V.W,*ASNV,\NV.\*.\ .
ШЖк
w
11. Sonli ketma-ketlik «„+i = 3 + 2ян rekkurent formula va ал f s5 shart
■vj
M*.%V
ass;?.
Л*Л*Л . *
AlUI MliTIK PllOtiUKSSIYA.
1. Arifinetik progressiva n — liaili formulas!.
!SS-.-.
SSSS*..
SWASSSS
'•.•.•.V.WV.V.V.V.'.
.•.•.•Л'/Л'Л'Л'Л'Л’
•.\4*.\W.\\N\\Vw*
..•.■.•.V.V.S4V.V.V
*
111
= 3 ;
= -7
=3-5*1. 2) fliitln + 2 ; 3) </„ =3h+7 ;
Чй]ШтV*5(5 -'7") ; 6) a. = 3 + 2(4« -1).
AVAVAV.'.NW.V. . V.\V.\\V.V.V.V .V.V.V.*
\>S4%V.y.V.VS*.V\Vh \V.V.\V.V.V.‘.‘.\ .v.v.v
.V. ■■■Х&Ш&.у-* ЩУ л
ГГГ^С C'tЧ /<Э П Q *
«]8ni; 2) «, = -4, d = 7 bo’lsa «24ni;
^ bo’lsa 4^,2 ni; 4) «, = 1, d = -2 bo’lsa «9ni;
ippifll d = 0,4 bo’lsa «31 ni; 6) «, = 6 , d = 1, .
.v. Ж
«! 5|0ff d = 2 bo’lsa «isni; 8) «, = 5,1, d = -0,3 bo’lsa «
«1 = -4,5, d = 1,3 bo’lsa «43 ni toping.
Arifmetik progressiyaning n - hadi formulasini yozing.
1 A 1 1 i a • ^ n n
1,2 bo’lsa «26 ni
ni;
5)
7)
9)
4.
1) 1, 6, 11, 16, ... ; 2) 25, 21, 17, 13,
3)-4, -6,-8, -10, ... ; 4) 1, -4, -9, -14, ... ;
5) - 8, -2, 4, 10, ... ; 6) 2, 14, 26, 38, 50, ... .
5. -22 soni 44, 38, 32, ... arifmetik progressiyaning hadi. Shu son
progressiyaning hadi bo’la oladimi va nechanchi hadi?
SOLIYEVX.X.
#44#
01/11/2015 у
***** у к НОВ *****
6. 12 soni -18, -15, -12, ... arifmetik progressiyaning hadi. Shu
son progressiyaning hadi bo’la oladimi va nechanchi hadi?
7. - 59 soni 1, -5, -11, ... arifmetik progressiyaning hadi. Shu son
progressiyaning hadi bo’la oladimi va nechanchi hadi? - 46 - chi?
8. Arifmetik progressiyada:
1) a, = 7, °i6 =67; 2) o, =-4, a, = 0; 3) o, = -2, o23 = -246 ;
4) a, = 2, a35= 104 ; 5) a, = -l, a,, =-31; 6) a, =9, <W= 144 ;
• • • • « « • • • • • • I I О • I
VAW.V.VV.W.NW.S • Л V
4 . • WA'.V.S'AV
vs- ■'■'vsvsv'
%
__ • • ■ •
7) a, = -3,5 , #8 = -22,5; 8) #, = 0, #28 = 729 ; 9) #, = jy|, = 55J||
‘.•.•A'.'.V.V.V.V.VAW.’.'.'.-A’.S. V.V.V.V.'
л fm A
10) #, = 1,4, a1 = 25,4 bo’lsa uning ayirmasini toping.
8. Arifmetik progressiyaning ayirmasi 1,5 ga teng. Agar:
1) a9 = 12 ; 2) «7 = -4 ; 3) #n = 23 ; 4) = 47 •
5) #16 = 28 ; 6) tf32 = -14,5 bo’lsa #!
9. л ning qanday qiymatlarida 15, 13, ,11,
ning hadlari manfiy bo’ladi.
10. Arifmetik progressiyada #i = ^lllig/% 0,5 bo’lsa, n ning qanday
qiymatlarida #„ < 2
Ж
ж,
11. Agar arifmetik progres;sifi|||fe-:% %,
SAs
Ш? щы
1) «. = 126, «„ = 14бЛ = -50 ;
щ
ssy.-
WW.VA NVAVA
♦WIAWAS.II
•AV.VAVA\V.
' vX::*S::x:x:x-.
3) o8 = -7, al0 = ||' j4M|B234r e10 = 256 ;
5) = -32 . al0 =/|§: ; -23,5, аю = -13,5 bo’lsa, uning
vSKvSSSSxvX WV
jW VA'AVJ.W.S'AS Wff
to’qqizinchi hadini va ayirmasini toping.
ЩШШШ: .
12. Erkin tushuvchijism bifinchi sekundda 4.9 m yo’l bosadi, keyingi har
bir sekundda esa oldingisidan 9,8 m yo’l bosadi. Tushayotgan jism
masofani bosib o’tadi?
1УHavoШгщаЩш olish yo’li bilan davolanishda birinchi kuni
XvX v. •лул'\'!у! ‘ly.j! *.s ч. ч*
dllijiiJlJffPminut davom etadi, keyingi har bir kunda uni 10 minutdan
osliifib boriladi. Vanna olish ko’pi bilan 1 soat 45 minut davom etishi
uchuiljp’fsatilgan tartibda havo vannasini olish uchun necha kun davom
etadi?
14. Arifmetik progressiyaning ayirmasini toping, uning o’ninchi va o’n
to’qqizinchi hadlarini toping.
1)8,
si 9I
4’ 2 ’
2)18, 12, 6,
3)1, 1 + л/з, 1 + 2л/3,
4)V2, V2-3, л/2-6,
#45#
01/И/2015 у
***** V KITOB *****
5)25, ... ; 6)л/5 + л/7, л/5+Зл/7, Т5 + 5л/7
15. л - hadining «„ = -5(2 -Зи) formulasi bilan berilgan ketma-ketlik
arifmetik progressiya bo’lishini isbotlang.
16. Agar arifmetik progressiyada:
1) = 8, d = 0,25 bo’Isa, <% ni toping.
_ ,1 , i
2) «i - -3 j, a bo’lsa, «25 ni toping.
Jt
"W
• V.\v.w>:.v
% P
:$> Xy.
V:>. W
V.V.*
• •
* >
3) «, = л/з , d = 2 bo’lsa, % ni toping.
а3 Л-4
4) a, - -Oy, a ~ — bo’lsa, «22 ni toping.
5) ax = 11} d = 2,5 bo’lsa, «9 ni toping.
, 1
6) «, = 15, d ~ 2 bo’lsa, «3i ni
17. Dam oluvchi shifokor tavsiyasiga amal qilib, birinchi kuni quyosh
nurida 5 minut toblandi, key ingyiqi: Fkunda esa toblanish 5 minutdan
oshirib bordi. Agar u toblanishni chorshahUkunidan boshlagan bo’lsa,
haftaning qaysi kuni uning§qu^Hda toblanishi 40 minutga teng bo’ladi?
18. Agar arifmetik progressiyada + «2 + a3 = 15 va «1 • a2 • «3 = 80
.%4v.\yX\s\y!sv!,.v!yX*.;. ‘v/XvXy
bo’lsa, uning birinchinalftMilvirmasini toping.
19. Agar arifmetik pfggressiyada «i + a2 + a3 = 0 va a\ + a\ + a\ - 50
bo’lsa, uning birinchi hadi.ya ayirmasini toping.
• \V.VA%y»v.S*A •»> *•*•*•*.*•*•*•*•*•*.*
20. Soat 1 da sofl! marta, 2 da 2 marta, ... , 12 da 12 marta bong uradi.
Soat mili navbatdaglhar soatning yarmini ko’rsatganda esa bir marta bong
uradi. Щsoat bir sutkada necha marta bong uradi?
2jT AfifMetik progressiyada «4 -a2 =4 va «7 = 14. Shu progressiya-
ning beshinchi hadini toping.
VA*.
уЛ*Л'.у.у.*ЛЧЛ Ww.s
22. Arifmetik progressiyada «2 = 12 va «5 = 3. Shu progressiyaning
o’ninchf hadini toping.
23. Arifmetik progressiya uchun quyidagi formulalardan qaysi biri
to’g’ri?
a„ -a.+d
1) al -2a2 +a3 =0 ; 2) «, = a3-a2 ; 3) n ~
24. Arifmetik progressiyada «2о =0 va «21 =-41 bo’lsa, ал ni toping.
25. Arifmetik progressiyada a2-a, =6 bo’lsa, «8 -a6 ni toping.
SOLI YE V X. X.
#46#
01/11/2015 у
***** у KITOB *****
26. Arifmetik progressiyada «2 = 9 va «26 = W5 bo’lsa, shu prog-ressiya
birinchi hadi va ayirmasi o’rta proporsional qiymatini toping.
27. Agar «, t a2,..., an sonlar arifmetik progressiya tashkil qilsa,
111 1
yig’indini toping.
+
+
4-... -f
a, •a.
a„ 'a
a, •a
a.
a
1 2 “3 “3 M4 ~/i-l
28. 4 ; 9 ; 14 ;... arifmetik progressiyaning sakkizinchi hadi to’rtinchi
hadidan nechtaga ortiq?
29. Arifmetik progressiyaning barcha hadlari natural sonlaAn iborat.
Agar я, = 3 va 20 < я3 < 22 bo’lsa, progressiyaning ayirmasini toping.
30. Arifmetik progressiyaning birinchi hadi 6 ga, oxirgi hadi 39 ga; teng.
Agar progressiyaning ayirmasi butun sondan iborat bo’lsa, oxirgi: haddan
oldingi hadlar sonini aniqlang.
31. 7 ; 10 ; 13 ;... arifmetik progressiyaning nechta hadimng har bin¬
ning qiymati 100 sonida katta, 200 sonidan kichik bo’ladi?
32. Arifmetik progressiyada «, - 3 va d - 2 bo’lsa,
ал - a2 + a3 - a4 + ... + a25 -a26 + «^ning qiymatini hisoblang.
,vXvX% X%<vXv!*.*.va'!v.va .
1 л. ^ ^ чШянШЬ
33. "T ; ~T,... arifmetik p!^gbs|iyaninfriechta hadi manfiy?
34. Arifmetik progressiyaning to’itinchi va o’n birinchi hadlari mos
ravishda 2 va 30 ga teng. Shu progrlpiyaning uchinchi va o’ninchi
hadlari yig’indisini|toping"
35. Kinoteatming birinqhi qatonia 21 ta o’rin bor. Har bir keyingi
qatorda o’rinlgsoni oiHingi qatordagidan 2 tadan ko’p. 40-qatorda nechta
o’rin bor?.^ Щ
:v:W:^$>№ss. ЩУ
я 2 + я 5 - <?3 =10 va «, + «6 = 17
bo’lsa, ll|j|g dl||jichi hadini toping.
a\ ~ 1 , «5 = 5 + X va я is = 10 + 3*
* V.y.y.V.4J.V.V,V.V^
‘ ЧХ|Х::;Х;>Ху.у\Х:.в
36. Agar
V.V.’.'.'.'.'A'.'.W*.
•.SSV.VSSSV.S'.SN
bo’lsa^ ^37 ni toping.
38. Arifmetik progressiyada «, = 1 , я5 = 5 + x va «,5 = 10 + 3*
bo’lsa, «39 ni toping.
39. Arifmetik progressiyaning ikkinchi hadi -7 ga, beshinchi va sak¬
kizinchi hadlaring ayirmasi - 6 ga teng. Shu progressiyaning nechanchi
hadi 9 ga teng bo’ladi?
2
#47#
01/11/2015 у
***** V KITOB *****
2. Arifmetik progressiva vossasi.
1. Arifmetik progressiya uchun #„ + ak - an_, + ak+l tenglik o’rinli
ekanligini isbotlang. Agar #7 + я8 = 45 bo’Isa +as ni toping.
&n+k ^ n—k
2. Arifmetik progressiya uchun a„ = ~ tenglik o’rinli ekanligini
isbotlang. Agar #10 + #30 = 246 bo’lsa #2о ni toping.
3. - 20 va 8 sonlari orasiga bitta sonni shunday qo’yinki, natijada
arifmetik progressiyaning ketma-ket uchta hadi hosil bolfllllfc>,4
4. Agar arifmetik progressiyada:
1) ai6 = 28, o18=34; 2)a16=-24, л„ =-6;
II
= 11
%
ip
a, о = m
Ft bo’lsa uning o’n
XsXsXvX-.
W
3) a\6 = “12 , л18 = 12 ; 4) #i6
yettinchi va birinchi hadlarini toping.
5. * ning qanday qiymatlarida sonlar arifmetik progressiyaning ketma-
ket hadlari bo’ladi?
Л’Л'Л'Л
1)3*,
x + 2
2x-l
:Шт
I’vXy’
ЙЗх
3) 5x , 3, x
Ik
v‘w
5x
4)3.v-7, a + 2, 5a-1 f| 5M1L ;|fc-,
6. Qyuidagi sonlar arifmetik progressiyaning ketma-ket uchta hadi
bo’ lishini ko’rsatinplk
1) sin(a + j3) ^ sin a cos ft , llin(a - /?) ;
* VV.V.,.,.V.V.V.%V.,iV.
• V.V.V.V.V.V.V.V.V.4
' Xs'vXvIv.wIv.
'x:x:x:x;:::::W:v::A
•.•.•.VAVAW
2) cos(or +
л-.y.y.s
Jpkcos a
••torn
_
, COS (a-/?);
4) sin 5or, sin 3or cos 2a , sin a
+ a.
И+1
rekkurent formula va #, = -2 va
й
щ
SWA’A'.V.VAV.
VASSW.VAWA
V.V.V.V.V.'.WA
‘.'.•.•.V.VWAWA.
SSV >AWAWA
SW.NWAW.SW. .-
shartlar bilan berilgan. Shu ketma-ketlikning yettinchi hadini
7. Arifmetik progressiyaning dastlabki 6 ta hadlari 7, #
2, *3
#4, *5,22
bo’lsa, #2 + *з + *4 + *5 ni hisoblang.
8. Ikkinchi, to’rtinchi va oltinchi hadlarining yig’indisi -18 ga teng
arifmetik progressiyaning to’rtinchi hadini toping.
9. Birinchi hadi 1 ga, o’n birinchi hadi 13 ga teng bo’lgan arifmetik
progressiyaning oltinchi hadini toping.
10. Ikkinchi hadi 5 ga sakkizinchi hadi 15 ga teng bo’lgan arifmetik
progressiyaning beshinchi hadini toping.
SOLI YE V X. X.
#48#
01/13/2015 у
***** у KITOB *****
11. To’rtta banderolni jo’natish haqi uchun jami 120 so’mlik har xil
pochta markasi kerak bo’ldi. Agar markalaming baholari arifmetik
progressiyani tashkil etib, eng qimmat marka ong arzonidan 3 marta
qimmat tursa, eng qimmatining bahosi necha so’m bo’ladi?
12. 0,(328); x va 0,(671) sonlari arifmetik progressiyani tashkil qiladi
x ning qiymatini toping.
13. m ning л/т-1 ; *j5m-1 va y/\2m + \ lar ko’rsatilgan tartibda
arifmetik progressiya tashkil qiladigan qiymatlari yig’indisinftoping.
wsv.v,
14. Uchta sonning o’rta arifmetigi 2,6 ga, birinchi son asaiyfga tenge
Agar keyingi har bir son avvalgisidan ayni bir songa farq
'ЛЧу.УЛЧУЛЧЧ'Л.’.'Л’Л'Л'АТ?
• w.y.y.'
sondan oldigisining ayirmasini toping.
15. O’suvchi arifmetik progressiyaning dastlabki uchta hadininjyi-
g’indisi 24 ga teng. Shu progressiyaning ikkinchi hadini toping.
AvlvSN'IwIy., ’ .vXv.'.V.
16. Arifmetik progressiyaning birinchi va to’rtinchi hadi yig’indisi
26 ga teng, ikkinchi hadi osa beshinchi hadidan 6 ga ko?p. Shu
progressiyaning uchinchi va beshinchi hadi yig’indisini toping.
.VAVVAV
.VAVAV
•УЛ'А'Л
3. Arifmetik progressiya n
Am«v. •№.
1. Agar arifmetik progressiyada:
1) a, = 3,
= 1.
.•l;Xvlv4
.•.V.V.’.'.V.V.'.V,*,
* ‘y.v.v.v.y.v.v.;. .v,v.y
тМШ&,
SWAV.VAV.
a = 51
n
%
3)«,
«, = 150
у
vX*XwAW>>A,Xw!,jl*I,X
* •AVW.VAW.^S,.^A,A
n =
hmph
уууУ
•A*
a.. = -56
71
a„ = 275
и = 18
n = 24
WAWA'AWAV
•A-V.V.VA*AV
■ •AWAVAV.' ■
4) «, = 5.
5) «з, = -4, a = 60 : 6) 1.5. =12,6, « = 70
Л .* «••••••••••
%.y.v\v
bo’Isa, unning dastlabki n ta hadining yig’indisini toping.
2. 7 dan 1 lllgaciia bo’l^n barcha natural sonlar yig’indisini toping
(llOhamyig^n
3. 1 dan 155 gacha bb’lgan barcha toq natural sonlar yig’indisini toping
A'.SVAVA
А'Л'А'ЛЛ
4. 1 dan 180 gacha bo lgan sonlardan 4 ga karrali barcha natural sonlar
yig mdisim toping.
•XyMv.
5. 1 datWOO gacha bo’lgan sonlardan 3 ga karrali barcha natural sonlar
yig’indisini toping.
6. 7 ga karrali bo’lgan barcha uch xonali sonlar yig’indisini toping .
7. Barcha ikki xonali, uch xonali sonlar yig’indisini toping.
8. Agar arifmetik progressiyada:
1) ="6, d = 4 ; 2) = 2,5, d = -2 ; 3) я, = 12 , d = 7 ;
#49#
01/11/2015 у
***** у KITOB *****
2) Agar n = 20 bo’lsa, 1, 8,
3) Agar n = 12 bo’lsa, -8,
4) Agar n = 18 bo’lsa, - 4 ,
4) o, =8, d = -5 ; 5) «, = 1, d = 25 ; 6) a, =-1,3, </ = 10
bo’lsa, uning dastlabki o’n sakkizta hadi yig’indisini toping.
9. 1) Agar « = 15 bo’lsa, 8, 14, 20, ;
15, ... ;
-13, -18, ... ;
-2, 0, ... ; arifmetikprogressiyaning
dastlabki n ta hadi yig’indisini toping.
10. Yig’indining qo’shiluvchilari arifmetik progressiyaning ketma-ket
hadlari bo’lsa, shu yig’indini toping.
1) 3 + 6 + 9 + ... + 318 ; 2) 120 + 80 + 40 + ... + (-160) Т'ЧШь,;
3) 21 + 28 + 35 + ... + 343 ; 4) -38 + (36) + (-34)
5) 90 + 80 + 70 +... + (—60) * 6) 12 + 22 + 32 + # + 242 .
11. Arifmetik progressiya n - hadining formufasi bilan berilgan.
Ч ч Ч ч. ..ллчч
1) a„ =Зи + 7 ; 2) a, =5n-2 ■ 3)a„ =
*****
4) an = 18 + In ; 5) an = 11« - 6 bo’lsa S60 va toping.
12. Ketma-ketlik a,l+] = an - 5 rekurrent formula va o, = 11 shart bilan
.WvavXsvvIvIv.'X*. .
AV.SW.V.VAWA*A4W.'A .
mm
mmm-
• *
ЛЧ’А’А*.
•V.V.V.N
VVA’.V
VA'.V
P
•Л,А‘Л‘.*А*>АЧ*.*Л,Л,.,Л,.*Л,Л'.
•.•.V.N’.VA*.'.'
- - .VAW.V.S
.y^\w.;.Wv.v.s\y.w
Vv!‘!v!vXv!\v.v.v’
•v.v.vaw.v.v
w.wv.v
чч?лЙг
Jlk
berilgan. Shu ketma-ketlikning dastlabki oWbeshta hadining yig’indi-sini
toping.
• Л • » • ft • • • • • • • • »
‘.WAWAV.-.V.*.*.
VA*A,A,.V.,.,.,.,A,A*.
13. Yig’indi 75 ga teng bo’lishi ucKun ЗШап boshlab nechta ketma-ket
natural sonni qo’
14.
y>:-
v.v/Av/awa*.
WA’A'A'A’A'A*
■ву’
w
4^
I .'A'i.'A'A'.VA Ami
1) о, = 10, пМ\ 4
•А'А'А'Л'Л'Л’Л
• • ' • •
•AASWA'A'A'.'.
ill
Г
3) «, - 78
2) *i
= 2*
3
n = 10
■AW.V.V.V
WAW
Sn -676 ; 4) o, = -12 , n = 20
s*=90l
S„ = 800
ч\\чч\\*
в V.V.4W.V.V.V.4
VA-A'.V.'.'.'/A'A
bo’lsa Щ». va ni pping.
.'.'.■.VA'.VA VA*A4S4S,a',.S *.v.v
A'A'AVV.VA ЧЧЧЧЧЧЧ’.ЧЧЧ*.* V.V
15. Agarar
266; 2) «и
4) «25 ='46 ; 5,3 =529 bo’lsa
16. Arifmetik progressiyada о
17. Arifmetik progressiyada oI2 + o25 ~ 73. S36 ni toping.
= 92, Su =22; 3)«15 =144
va d ni toping,
з + a9 = 18 . S,, ni toping.
Sl5 = 495
18. Arifmetik progressiyada o7 + a\6 = 82 • S22 ni toping.
19. Arifmetik progressiyada o16 +a31 = 107 . S52 ni toping.
20. Arifmetik progressiyada o5 + a22 = 25 . S26 ni toping.
SOLIYE V X. X.
#50#
Ol/H/2015 у
***** VKITOB *****
(110 ham yig’indiga kiradt).
3. 1 dan 155 gacha bo’lgan barcha toq natural sonlar yig’indisini toping
(155 ham yig’indiga kiradi).
4. 1 dan 180 gacha bo’lgan sonlardan 4 ga karrali barcha natural sonlar
yig’indisini toping.
5. 1 dan 100 gacha bo’lgan sonlardan 3 ga karrali barcha natural sonlar
yig’indisini toping.
6. 7 ga karrali bo’lgan barcha uch xonali sonlar yig’indisini toping .
7. Barcha ikki xonali, uch xonali sonlar yig’indisini toping \
8. Agar arifmctik progressiyada:
1) *, = -6, J = 4 ; 2) a, =2,5, d = -2 ; 3) a, = 12 , *</«'? ;
4) a, = 8, </ = -5 ; 5) в| * 1, </ = 25 ; to ax = -1,3, ф 10
bo’lsa, uning dastlabki o’n sakkizta hadi yig’indisini toping.
9. 1) Agar л = 15 bo’lsa, 8, 14, 20,
2) Agar л = 20 bo’lsa, 1, 8, 15, ...a
3) Agar w = 12 bo’lsa. -8, -13, -18, ; v
4) Agar n = 18 bo’lsa, - 4 , -2,0,^; arifmetik progressiyanmg
dastlabki n ta hadi yig’indisini toping. $ к
10. Yig’indining qo’shiluvchilari arifmetik progressiyaning ketma-ket
hadlari bo’lsa. shu yig’indim toping. jL'
1) 3 + 6 + 9 + ... + 318 ; 2) 120 + 80+ 40+ ... + (-160) ;
3) 21 + 28 + 35 +,.. + 343 ; 4) - 38 + (36) + (-34) +... 12 ;
5) TO + 80 + TO+^&NS0) Щ 12 + 22 + 32 + ..+242 .
11. Arifmetik progressiya n - hadining formulasi bilan berilgan.
1) ". |) л- = 5'1"2 i 3) = 2(7n-5)+l ;
4) a. i= 18+ 7/i ;~5) aH = 1 l/i-б bo’lsa Sva Sn ni toping.
12. Ketma-kctlik = ая- 5 rekurrcnt formula va ox = П shart bilan
berilgan. Shu ketma-ketlikning dastlabki o’n beshta hadining yig’indi¬
sini toping.
13. Yig’indi 75 ga teng bo’lishi uchun 3 dan boshlab nechta ketma-ket
natural sonni qo’shish kerak?
14. Agar arifmetik progressiyada:
l)a, = 10, /1 = 14, S. = 1050 ; 2) ". = 2 J, л = Ю, 5- = 90^.
3) o, = 78 , n = 26 , S, = 676 ; 4) a, = -12 , n = 20, S. = 800
bo’lsa va d ni toping.
SOLIYEV X. X»
# 51 #
02/03/2015 У
***** V KITOB *****
«I
/X;
illfe
yi-
31. Ikkinchi va o’n to’qqizinchi hadlarining yig’indisi 12 ga teng
bo’lgan arifmetik progressiyaning dastlabki yigirmata hadining
yig’indisini toping.
32. 100 dan katta bo’lmagan 3 ga karrali barcha natural sonlaming
yig’indisini toping.
33. Arifmetik progressiyada a2 = Ю va ci5 = 22. Shu progressiya¬
ning dastlabki sakkizta hadining yig’indisini toping.
34. Arifmetik progressiyada a3+a5 =\2, S7 ni toping.
35. Arifmetik progressiyada я4 +a6 = 10. S9 ni toping.
36. Hadlari x„ = 4/2 + 5 formulabilan berilgan ketma-ketlikning dast¬
labki o’ttizta hadi yig’indisini toping.
37. (*„) arifmetik progressiyaning dastlabki n ta hadi yig’indisi
120 ga teng. Agar *3 + *„_2 = 30 bo’Isa, yig’indida nechta had qatnashgan?
38. 150 dan katta bo’lmagan 7 karrali barcha
g’indisini toping. jfL \ Jps
39. G’o’la shaklidagi to’sinlar ustmaliist taxlangan. Birinchi taxlamda
10 ta, ikkinchi taxlamada 9 ta, :ya h,k oxifgi taxlamda 1 ta to’sin bor.
Taxlamda nechta to’sin bor? lj||kll
40. a„ = 4w - 2 formula bilan berilgan ketma-ketlikning dastlabki
50 ta hadining yig’indisini toping. l|Pr
41. 100 dan katta bo’Imagan 4|||;karrali barcha natural sonlaming
yig’indisini toping,
42. Dastlabki;eftita hadining yig’ indisi -266 ga, dastlabki sakkizta
hadining yig’indisi - 312 ga va hadlarining ayirmasini -2 ga teng
etik proressiyaning birinchi hadini toping.
43. Qu^l^rfistma-tist taxlangan. Birinchi qatlamda 11 ta, ikkinchi qatlamda
10 ta% i||:oxirgi qatlamda 1 ta quvur bor. Taxlamda nechta quvur bor?
44|||Щап katta bo’lmagan 6 ga karrali barcha natural sonlaming
yig’inciismi toping.
45. Hadlari bn =3/7-1 formula bilan berilgan ketma-ketlikning dast¬
labki 60 ta hadining yig’indisini toping.
46. Arifmetik progressiyaning dastlabki 16 ta hadlari yig’indisi 840
ga, oxirgi hadi (tf16) 105 ga teng. Shu progressiyaning ayirmasini toping.
47. Arifmetik progressiyada S20 -S19 = -30 vacf = -4 bo’lsa,
ning qiymatini toping.
2
48. Arifmetik progressiyada dastlabki n ta hadi yig’indisi Sn = n
bo’
SOLIYEV X. X.
#52#
01/11/2015 у
***** V KITOB *****
bo’lsa, uning o’ninchi hadini toping.
49. Arifmetik progressiyaning uchinchi hadi 8 ga, to’rtinchi hadi 5 ga va
dastlabki bir nechta hadlari yig’indisi 28 ga teng. Yig’indida nechta had
qatnashgan?
50. Agar arifmetik progressiyada Sn - S„_, = 52 va Sn+l - Sn = 64
bo’lsa, uning hadlari ayirmasini qanchaga teng bo’ladi?
51. Arifmetik progressiyada a2 +al9 = 40. Shu progressiyaninng dast¬
labki 20 ta hadlari yig’indisini toping.
52. Arifmetik progressiyaning uchinchi va beshinchi hadi mos ravish-da
NVJ.JAWWJw.V.ViViViV.ViVi
11 va 19 ga teng bo’lsa, uning dastlabki o’nta hadlarining yig’indisi
qanchaga teng bo’ladi?
53. (a„) arifmetik progressiyada cix = 3, aM = 57 bo’lsa, progressiya¬
ning dastlabki 60 ta hadi yig’indisini toping,
54. Tenglikni qanoatlantiruvchi natural son"'7Pllllpbg,<
1 2 3 N
+
+
+... +
= IOOtV
AWAWAV
A^'A'AWa
AW.SYAHSSWA
a:-s*a%ya>s:a:.s\v
■s.
%
Щш
Ш
100 100 100 100
■y.svA;.vy.y.y.;.v.sj.v^ *.v
55. Arifmetik progressiyaning hadlari 19 ta. Uning o’rta hadi 21 ga
'vX*. •
teng. Shu progressiyaning hadlari yig’indM|||toping.
56. Agar a2+a4 +a6 + an_2 + a„+4 =42 bo’lsa,
ai, a2, a2n arifmetik progressiyaning hadlar sonini toping.
57. 5 ga bo’lgandadoldiq 1 chiqadigan dastlabki 20 ta sonning yig’in-disii
9 A/Xwv/ava'X*. л*л*а*л,л,а,а,‘
toping.
58. Arifmetik progressiyaning o’n uchinchi hadi 5 ga teng. Uning dast¬
labki 25 ta hadlarining^yiflmdisini toping.
VA4V.V
F
59. а \ЗЖШ
•.•.v.sv.w.s'.sy
4v.v.-.4\v.s;.s<
1I1E
•' 1
VAWA'AWAV
П VA'A’A'A'A'A*.!-
fitma-ketlikning dastlabki 10 ta hadi yig’indisi
255 ga l^g^lKlngiqiVmatini toping.
60Apftlllik progressiyaning dastlabki n ta hadi yig’indisi 91 ga teng.
jfl| * ^ #
Agar ^|= 9 va % ~ch ~ 20 ekanligi ma’lum bo’lsa, n ni toping.
AVAVA44VA*A,AV.,A\V.VA,.'.,A
61:100 dan ortiq bo’lmagan 3 ga karrali natural sonlar yig’indisini
♦УА.ч
62. Barcha ikki xonali sonlar yig’indisi qanday raqam bilan tugaydi?
63. Arifmetik progressiya 26 ta haddan iborat. Agar a6 = -0,25 va
a2l = -0,5 bo’lsa, uning hadlari yig’indisini toping.
64. Arifmetik progressiyaning dastlabki to’rtta hadi yig’indisi 124 ga,
oxirgi to’rttasiniki 156 ga teng. Progressiyaning hadlari yig’indisi 350
ga teng. Progressiyaning nechta hadi bor?
65. Natural sonlardan qatori har biri sonning kvadrati bilan tugaydigan
SOLIYEVX.X.
#53#
01/1V2015 у
***** V KITOB *****
quyidagi qismlarga ajratilgan: 1, (2, 3, 4), (5, 6, 7, 8, 9), (10, 11, 12, 13,
14, 15, 16),... 10-qismidagi sonlar yig’indisini toping.
66. 25 ta ketma-ket natural sonning yig’indisi 1000 ga teng. Bu sonlar-
ning kichigi nechaga teng bo’ladi?
67. Arifmetik progressiyaning dastlabki 13 ta hadi yig’indisi 104 ga
teng. Yettinchi hadining kvadratini toping.
68. Arifmetik progressiyaning dast-labki nechta hadini olmaylik ular-ning
yig’indisi hadlar soni kvadratining uchlanganiga teng. Shu
progressiyaning yettinchi hadini toping.
69. Arifmetik progressiyaning beshinchi hadi 6 ga teng.
tlabki to’qqizta hadi yig’indisini toping.
70. 1 dan 75 gacha bo’lgan toq sonlar yig’indisi qanday raqam bilan tugaydi.
71. O’zidan oldingi barcha natural sonlar yig’mdisining qismiga
*Авднайййййшвй&. ‘
■'ly.v.v.v.y.v.v*
Xv!v!v.v!v.
teng bo’lgan natural sonni toping.
72. Arifmetik progressiyaning dastlabki?akkizta hadi yig’indisi 32 ga,
dastlabki yigirmata hadining yig’indisi 200 ga teng. Progressiyaning dastlabki
28 ta hadining yig’indisini toping!
73. Arifmetik progressiyaningllirin^hi va ®qqizinchi hadlari yig’indisi
S'™;;-. ^ ^
64 ga teng. Shu progressiya|iin||i|s|Ilbki to’qqizta hadlari yig’indisi va
beshinchi hadi ayirmasini toping.
74.5 va 1 sonlari oras i ga shttisonlar bilan arifmetik progressiya tashkil
etadigan bir nechta joylashtirildi. Agar bu sonlarning yig’indisi 33
ga teng bo’lsa,:nechta hadjoylashtirilgan?
jVlwl'w. \ул\ул*л*.ул
75. у, Зу + $||Щ + 10 arifmetik progressiyaning dastlabki 8 ta hadi
yig’indisi 396 gaf^hg. lining qiymatini toping.
76. Aga^soaill da bir marta 2 da 2 marta va 12 da 12 marta zang ursa,
'«I-M-X’X-’v M'l*
bir sutkada necha marta zang uradi?
77. Arifmetik;progressiyaning dastlabki uchta hadi yig’indisi 66 ga
ikkinchi Ш uchinchi hadlarining ko’paytmasi 528 ga teng. Progressiya¬
ning birinchi hadini toping.
78. 2 va 65 sonlari orasiga 20 ta shunday son qo’yilganki natijada hosil
bo’lgan ketma-ketlik arifmetik progressiyani tashkil otadi. Shu prog-
ressiya hadlarining o’rta arifmetigini toping.
79. 1 dan 50 gacha bo’lgan toq sonlar yig’indisining kvadrat ildizini
toping.
80. 1, 8, 22, 43 sonlar ketma-ketligi shunday xususiyatga egaki ikkita qo’shni
hadlarining ayirmasi 7, 14, 21,... arifmetik progressiyani tashkil etadi.
Berilgan ketmaketlikning nechanchi hadi 35351 ga teng bo’ladi?
SOLIYEVX.X.
#54#
01/11/2015 у
***** VKITOB *****
81. Arifmetik progressiya uchun quyidagi formulalaming qaysilari
to’g’ri?
a..-ax+d_ _ c ax+(n-\)d
1) #, + #„ = #3 + #„_2 ; 2)
3)
S„ =
n
n ' ' 2
82. Arifmetik progressiyada 20 ta had bor. Juft nomerli hadlar yig’in-disi
250 ga, toq nomerli hadlaming yig’indisi 220 ga teng. Progressiya-ning
birini hadi va ayirmasini toping.
83. Arifmetik progressiya uchun axl = 2 ga teng bo’Isa, S2df,Sl2 ni
toping.
2 nr\2 . no2 m2
A m
iiiliifc,.
84. Yig’indini hisoblang. 100' -99' + 98' -97' +... + 21 *i||||
в
Ш
85. Hisoblang. 1002 -972 +962 -932 +922 -892 + ... + 42 -1
.v.v.v.
■ЛЧУЛЧУИ/Л’.'
* WMVAV
йЩ
ХХямОС
%
у
w
1
.•.•.'.•.•.’.•.•л*.*
i 8 ta hadi
86. O’zidan oldin kelgan barcha toq natural sonlar yig’indisining — qismiga
teng bo’lgan natural sonni toping.
87. Ushbu -y/s , - y[2,...
yig’indisini toping. jlllt*,. \ w
88. Ketma-ket kelgan oltita natumi sbhnj|lg yig’indisi 435 ga teng.
Shu sonlarning ong kichigini tqpmg|k
_ ... УЛЧУЛЧЧЧУЛУ ЧУЛЧУЛ A
89. Ketma-ket kelgan etti$|patur|l|sonning o’rta arifmetigi nimaga teng?
90. Arifmetik progress!#,bifinchi o’nta hadining yig’indisi 140 ga teng
bo’Isa, a2 +a9 ni
УЛЧ^.УЛЧУЛУЛЛ
VW.V.VAW.y.V.SV.'A . .
УЛЧЧУЛЧУЛ'Л’.уЛУЛЧ'Л .
91. Arifmetik progi^i^aning qchinchi, yettinchi, o’n to’rtinchi va o’n
sakkizinchi hadlarinihjlli|’indisi 48 ga teng. Bu progressiyaning
dastlabki 20 te f|di yig’indisini toping.
92. 4, 7|:-1|Bl5plfarining o’rta arifmetik qiymatini toping.
XvX 'vlvIyXy/.v!*. XsivXwX’Aw.
Vggggt ^/ЛУ.У.УЛУ.УЛ •AW.y.y.V.SV
93. Arifmetik progressiyada ax = -3 va d = 5 bo’lsa, Sl5 -Sl4
pfogressiya hadlari uchun ushbu tenglik o’rinli bo’lsa,
#1 + С!з + ... + #21 = #2 ^4 "t" ••• + ^20 13
■Лу.У.УЛ ,ViV
95. Arifmetik progressiyada #, = 0 va d = 3 bo’lsa,
#3 + #6 + #9 +... + #33 ning qiymatini toping.
96. Arifmetik progressiya hadlari uchun
#, +#3 + ... + #19 = #2 +#4 +... + #20 +Ю tenglik o’rinli bo’lsa, arifmetik
progressiyaning ayirmasini toping.
# +2#+ 3#+ ... + «# 3#
97. Soddalashtiring.
n -2n-3
2 (и - 3)
SOLI YEV X. X.
#55#
Ol/H/2015 у
♦XvXy.
хх
ш.
***** у KITOB *****
98. Sakkizta ketma-ket kelgan natural sonlaming yig’indisi 700 ga
teng. Shu sonlardan eng kichigini toping.
A) 78 V) 84 S) 82 D) 80 E) 86
99. Tenglamani yeching. (x + l)+(x + 4)+(x + 7) + ... + (* +28) = 155
100. Bir xil raqamlardan iborat ikki xonali sonlar yig’indisini toping.
101. 1 dan 75 gacha bo’lgan natural sonlardan kvadratini 3 ga bo’l-
ganda 1 qoldiq qoladigan sonlar yig’indisini toping.
102. 9 ga bo’lganda, qoldig’i 4 ga teng bo’ladigan barcha ikki xonali
sonlarning yig’indisini toping.
103. 8 ga karrali barcha uch xonali sonlaming yig’indisiii^*^
1
104. O’zidan oldingi toq natural sonlar yig’indisining ~ qismiga
О
teng bo’lgan natural sonni toping.
105. Dastlabki mingta natural sonlarning o^MlgifaietigiMloping.
a„) ketma-ketlikning dastlabki лrta hadining yig’indisi
2 •*• '•■•"•XvX
S„ = 11 - 4л fomula bo’yicha hisoblanadi. + filing qiymatini
tooins
v v К *** . • •.•л-л-л-л-л*л-л*л'л
J vX'X'XvXvX v.v.y.wv.v.v.v.v.*.
Xv. *• ;.у.\;л;.у.;.ул;
107. Arifmetik progressiya hadlari 60 ta. Uningjuft o’rinda turgan
hadlari yig’indisi toq о’rin# turgan hadlari yig’indisidan 15 ga ko’p.
Progressiyaning to’rtincM Htdi 4Д§§| teng. Progressiyaning hadlari
yig’indisini toping. ;i
108. 15 ta haddan iborat arifmetik progressiyaning sakkizinchi hadi
18 ga teng. Shu progressiyaning hadlari yig’indisini toping.
109. Agar arifmetik progressiyaning dastlabki n ta hadining yig’indisi
bilan topilsa, uning umumiy hadi qanday
\N\s\v*v\y*\\y.y.v.
s„ =
п
*х-х*.
Ip 0. Aifrnetik progressiyaning oltinchi hadi 10 ga, dastlabki 16 ta
Хччу1уХ\Счччччч\у1у.\'.уХ\*Х\
hadining yig’indisi 200 ga teng. Bu progressiyaning o’n ikkinchi hadini
topin&Jf
111. m + 2- 8 + 3-12 + ... + 20-80yig’indidaharbirqo’shiluvchining
ikkinchi ko’paytuvchisi bittadan kamaytirilsa, bu yig’indi qanchaga
kamayadi?
112. 1-4 + 2-8 + 3-12 + ... + 30-120 yig’indida har bir qo’shiluvchining
ikkinchi ko’paytuvchisi bittadan kamaytirilsa, bu yig’indi qanchaga
kamayadi?
113. 1-4 + 2-6 + 3-8 + ... + 10-22 yig’indining har bir hadidagi ikkinchi
ko’paytuvchi 3 ta kamaytirilsa, yig’indi qanchaga kamayadi?
SOLIYEVX.X.
#56#
01/1V2015 у
***** V KITOB *****
114. Arifmetik progressiyada = 56 bo’Isa, uning dastlabki 19 ta
hadlari yig’indisini toping.
115. Soddalashtiring.
a+ 2a + 3a + ...na (
n2 -2n-3
slab -
ab
V
a +
л[аЬ
2^[аЬ - b)
a-b
116. Agar arifmetik progressiyada a, +a2 +a3 +.. + al6 +ал1 =136
bo’Isa, a6+a\2 ni hisoblang.
117. 7 ga bo’lganda, qoldig’i 2 ga teng bo’ladigan barch^lkki xonali |1|
sonlaming yig’indisini toping;.
118.21 ta hadining yig’indisi 546 ga teng bo’lgan arifmetik progres-
siyaning o’n birinchi hadini toping.
119. Tenglamaning ildizi 10 dan nechta kamTg
Jt-l x-2 jc-3 1
+ + + ... + — = 4
xxx x M
■;ll
^ %
.Л4
шш
> *» •% • в а ■ в к Л • а Л • • • • •*. »Ч Л
V.NW.V.V.V.V.V.V.V.V.'.'.'.'.V.
• W.V.vX'AWAW.V.V.V.SV
WW4V.VA:,W
120. Sn arifmetik progressiyaning dastlabki n ta hadi yig’indisi bo’lsa,
iSj - 3S4 + 353 — S2
nmg
121. 10 ; 15 ; 20 ;... arifmetik progressiyffiffig dastlabki nechta hadi¬
ning yig’indisi 2475 ga tepi:bdl|||i|
'•Ms'IvlsvMvX'.-
!>мл\ул:л:.
122. Dastlabki n ta hadinin| yig’indisi Sn =2n -3n formula
W"
bo’yicha hisoblanadi|anlfiffietik progressiyaning ayirmasini toping.
‘ s^vSxWXvXvbJ—
1. Geomttik Isiya n - hadi formulas!.
11111111Ш
4) КШЛ = 5
W"
2)6, =-5,? = 3
5 )b.=-\,q = -
3) *, = 1,9 = 4 ;
6) *, = 8,9 = 1;
1
7) *, = 15, 9 - - — bo’lsa dastlabki oltita hadini yozing.
2. и - hadining formulasi bilan berilgan quyidagi ketma-ketlik geometrik
progressiya bo’lishini isbotlang.
1) *» = 3'2“ ; 2) *„ = 5"*2 ; 3) *„ = {l)‘
3. Geometrik progressiyada:
-2
4) *„=-3-4
и—1
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_ 2
= 15 va 4 ~ ~ bo’lsa b4 ni;
1) b} = 2 va q = 10 bo’lsa b5 ni; 2) \ =
1
3) bx = 8 va 4 - — bo’lsa b% ni; 4) bx
5) bx = -3 va Я - ~ bo’lsa b7 ni; 6) bx = 1 va q = 3 bo’lsa b6 ni;
= 5 bo’lsa £8 ni; 8) bx = 9 va q = 6 bo’lsa 64 ni;
= 2 bo’lsa ni toping.
7)
9) bx
12 va <7
14 va q
4. Geometrik progressiya n - hadining formulasini topingj
1) 4, 12, 36,
1
2) 3, 1, j
3) 4, -1
4) 3, -4,
16
5) 2, 10, 50, ... ;
4
ЖЩ'Л’Л .
4 ’
2 4
11 — —
* 3 ’ 95
'••Ж
>Vt
• W
9 iy.
w
r
. .V.4V.
Х\\‘л\\Х
5. Geometrik progressiyada tagiga chizi 1 gan hadipmg>nomerini toping.
1)6, 12, 24, ...,192
3)625, 125, 25, ... ,
2)4, 12,
VW.V.\,.\W.%\4%4Nv,.V
A\V.Vt\W.V.4*.VA\V
- - A. .v.v.v.,/.v.v.v
V.1
>X<« •
1
Mi,
25
A
.128,...
l:*..
5)2, 6, 18, , 486,
.'.VV.V.V 'A’lV.S'XdtvWN .v.v.*.*.
W • ■УЛ V.'.S'.W.y.V.V AVAW
•л*. .V.V.V
34992 .
6.
о-
V.'.V.V.*.V.V.*A W«V
• 1Ш <r
1)^ = 2 va b5 = 1 |у-128 va b7=- 2 ■
3) ^=3 va b7 = 2187 ; 4) l|P 250 va b4 =-2 ;
>;.v ^\\vXsy.VA %v
5) bx =3 va 65 =1875 6) ^ = -5 va b7 = -320 bo’lsa uning
maxraj ini;
7. 2, 6, 18, Щ geometrik progressiya berilgan.
1) shu progressiyaning sakkizinchi hadini hisoblang ;
2} ketma-ketlikning 162 ga teng had nomerini toping.
8.Spimusbat hadli geometrik progressiyada:
rr 11
1) b6 va 68 = 81 ; 2) b6 = 16 va &8 = 64 ; 3) b6 = 1 va 68 = 25 ;
4) b6 = 9 va = 3 bo’lsa uning yettinchi hadini va maxraj ini toping.
9. Agar geometrik progressiyada:
1) b4 = 9 va b6
4) 64 = 5 va b6
36 ; 2) 64 = 18 va b6 = 8 ; 3) b4 = 20 va b6 = 5
8 bo’lsa uning beshinchi va birinchi hadlarini topi;
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10. Mijoz bankka 2002-yilning 1-yanvar kuni 5.000.000 so’m pul qo’ydi.
Agar bank yilliga 30 % i miqdorida daromad bersa, mijozning puli 2006-
yilning 1-yanvariga borib qancha bo’ladi?
11. Geometrik progressiyaning maxrajini toping va oltinchi va o’ninchi
hadlarini toping.
1)3,
!, 3,
2)
I _I __L
4’ 8’ 16’
-5V2,
3)3, л/3, 1, ... ; 4)5, -5V2, 10,
12. Geometrik progressiyaning n - hadining formulasini yozing.
1) - 2, 4, -8, ... ; 2)-0,5, 1, -
3) 16, 8, 4, ... ; 4) л/з , -n/6 ,
13. Agar geometrik progressiyada:
1) *i=2, q = 2, n = 6 ; 2) A, = 12 ,
лЧ^Йч
3)
= -
1 8
q = 5 , n = 4
5) *i = 25, <7 = -5, и = 6;
7)*,=1, ?= 4, « = 5
14. Geometrik progressiyadai
, F '"iiliii,
?i wsk - ' —:
IP
0,25 , и = 4
wy
••••••
,v.%
•AWAVAWAV.
•y.V.V.V.'.SV.'.SV.W.WSSS'.SV.*.*.*
L a #em. . .4
1) agar b\-T , q - :3 bo’Isa, % m toping.
2) agar bx = !%j|g V|jplsa, 67 ni toping.
1
2 bo’Isa, bw ni toping.
3) agar§
J|
ШЖЖЬ* Ш
Я = ~ bo’Isa, b6 ni toping.
15. Nolgapeng bo’lmagan x, у, z sonlar kshrsatilgan tartibda isho-
o’zgarufbhi geometrik progressiyani x + y ; y + z ; z + x sonlar
esa arifmetik progressiyani tashkil otadi. Geometrik progressiyaning
maxrajini toping.
16. Geometrik progressiyaning dastlabki 6 ta hadi 2 , b2 , Ьъ , b4 , b5
va486 bo’lsa, b4 +b5 ni hisoblang.
17. Quyidagi ketma-ketliklardan qaysilari geometrik progressiyani
tashkil etadi?
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/
1)я„=2дг"; 2) a„ =ax" + \ ; 3)
K =
3V
5
4)Й„=Г2”
5) a„ = 3'2“" + 5 ; 6)
*„ =
/
V
sin 60'
lY
3, •
18. Nechanchi haddan boshlab -8 ; 4 ;-2 ;geometrik progressiya
hadlarining absalyut qiymati 0,001 dan kichik bo’ladi.
19. 64 ; 32 ; 16 ... geometrik progressiyaning to’qqizinchi ha|i oltinchi
hadidan nechtaga kam.
20. Geometrik progressiyaning maxraji 0,5 gatengbo’
bl-(b2y>-br(biy'-...-bll-(buy[ ning qiymatini hisoblang.
21. Geometrik progressiya hadlari uchun
b, • b., •...• bt, = b* • b. •...• b
•x-x-x-::-
•V:. "W
IF
■bu -128 tenglik o’riilfelsa, 6||i toping.
i\/Q Ь 1 ЯГ1 11 nVl11 n
1 ~3 ••• ^13 4
22. Geometrik progressiya hadlari uchun
b\ *b3 з *128 = b2-b4 ‘’bX4 tengjif%’rinli bo’I|a, progressiya
maxrajini toping.
23. Agar geometrik progressiyaning lastlabki 4 ta hadiga mos ra-
vishda 1 ; 1 ; 4 va 13 sonlarini qo’shsak, ular arifmetik progressiyani
tashkil etadi. Geometrik progressiyaning maxraj ini toping.
24. Yig’indisi 35 ga tenflbo'lgan uchtason o’suvchi geometrik
progressiyaning dastlalllPliEfa hadlaridir. Agar shu sonlardan
mos ravishda 2 ; 2:!!!!1Жоп1 arini avrilsa. hosil bo’lgan sonlar arifmetik
progressiyaningketmallilhadilri bo’ladi. Arifmetik progressiyaning
dastlabki 10 ta hadining yig’indisini toping.
25. (b„ I^la^tnMlIrogressiyada b4-b2 = 24 va b2 +b3 = 6 bo’Isa,
b,
26. C = > 0) sonlar ketma-ketligining umumiy hadi bo’Iib,
VAWA\\WAS%WAWW>JWA * V. .V,V. . .%y *
С2'<^:~Ц bo’lsa, a nimagateng?
/ \
27. <% = a- k"~5 (a> 0) sonlar ketma-ketligining umumiy hadi bo’Iib,
'•Xv'
w
c2-c8 =36 bo’lsa, a nimaga teng?
28. Ikkinchi hadi 6 ga teng birinchi uchta hadining yig’indisi 26 ga
teng o’suvchi geometrik progressiyaning uchinchi va birinchi hadlari
ayirmasini toping.
29. O’suvchi geometrik progressiyaning birinchi hadi 3 ga, yettinchi
va to’rtinchi hadlarining ayirmasi 168 ga teng. Shu progressiyaning maxrajini
toping.
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2. Geometrik progressiya xossasi.
1. 4 va 9 sonlari orasiga bitta musbat sonni shunday qo’yinki, natijada
geometrik progressiyaning ketma-ket uchta hadi hosil bo’lsin.
2. Geometrik progressiya uchun b2n = bn+k • bn_k tenglik o’rinli ekanligini
isbotlang. Agar b3 -bn = 225 bo’lsa bn ni hisoblang.
3. Geometrik progressiya uchun bn -bk = bn+l -bk_, tenglik o’rinli ekanligini
isbotlang. Agar b3-b5 = 72 bo’lsa bx ■b1 ni hisoblang.
4. Agar ketma-ketlik n - hadining formulasi berilgan bb«iI|Ulliksil
kamayuvchi geometrik progressiya bo’la oladimi?
10 . . 25
1) b. = 5
/! +1
2) b„ = (-4)
w+2
3)
К =
m
bn = ~
r
7n J n 0/7-1
5. x ning qanday qiymatlarida 0,(16); л: ; 0,(25) sonlar illbralari
almashinuvchi geometrik progressiyaning ketmk-ket hadlari bo’ladi?
6. Geometrik progressiyada uchinchi va yettinchi hadlarining ko’payt-
masi 144 ga teng. Uning beshinchi hadini toping.
ч®у
7. Geometrik progressiyada Ь^ё3 kb4 =216 bo’lsa, uning uchinchi
hadini toping.
8. b3 •b4 •b5 =64
.
■ -.V.V.VV
.VAWAWAy
AWAWAW.NV
Sags *
M. v ж
ga teng bo’lgan geombtrik progressiyaning to’rtinchi hadini
toping.
9. Agar geometri
b»• be ni
ж
w
Фх + bg = 5 va bx +b9 =17 bo’lsa,
6
1
1
WAW.V.VA'.V
111,
л;. BSfimw
_ v _
■.■.•.■.‘.•.•.’.•.•.•..•.•Л'Л'Л’ЛЛЧЧЧЧ
\SSV.S‘.44SSS4S‘.^^4>V\N4V.y.V
Ч’Л'Л’Л'Л'ЛЧ'.-Л'Л'Л’Л'ЛЛ'Л'Л’Л
10. т уа:|Щ|:§ЬШйг or||iga shunday uchta musbat sonni joylashti-
.V.V. WAW-VAV/.S1. ^.V.V.V.y.y.y.S
у.ул\ w^\y.y.y.\\s\ улулу.улу
ringki, qptyasida: geometrik progressiya hosil bo’lsin. O’sha qo’yilgan
uchta sonnipg yig’indisini toping.
i musbat bo’lgan geometrik progressiyaning birinchi
haiiPgi||eshinchi hadi 18 ga teng. Shu progressiyaning beshinchi va
ayirmasini toping.
12. Hadlari musbat bo’lgan geometrik progressiyaning birinchi va
uchinchi hadi ko’paytmasi 4 ga, uchinchi va beshinchisiniki esa 64 ga
teng. Progressiyaning ikkinchi, to’rtinchi va oltinchi hadlari yig’indi-
sini toping.
13. 2 ; b2 va b3 sonlari o’suvchi geometrik progressiyaning dastlabki
uchta hadidan iborat. Agar bu progressiyaning ikkinchi hadiga 4 qo’-
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shilsa hosil bo’lgan sonlar arifmetik progressiyaning dastlabki uchta
hadini tashkil etadi. Berilgan progressiyaning maxrajini toping.
14. Kamayuvchi geometrik progressiya tashkil atuvchi uchta sondan
uchinchisi 18 ga Bu son o’miga 10 soni olinsa, uchta son arifmetik
progressiya tashkil etadi. Birinchi sonni toping.
15. 3 va 19683 sonlari o’rtasida 7 ta shunday musbat sonlar joylash-
tirinki, hosil bo’lgan to’qqizta son geometrik progressiya tashkil etadi.
5-o’rinda turgan son nechaga teng?
16. Agar 1 ; V? ; 3V? + 4 sonlari geometrik progressiyaningjketma-^
ket hadlari bo’Isa, у ni toping.
17. a, b, c,
mmm
fSL
4^#
tashkil etadi. (a - c)2 +(b-c)2 + (b-d)2 - {a -d)2 ni soddalashtiring.
3. Geometrik progressiya n
1. Agar geometrik progressiyada:
mm
.-.v.vXy:*..
л'^л
VAV.'A
.•AV.V.V.
,V.VA\V,S
•.VAV.N*
VNV4V
xf
ivKv
•XSv.w
.'.'.vAW
1) = 2. <7 = 5, n = 7 ; 2) М*УЗШ*| -, n = 5
1
* ‘•XvXsvXvXvl;
Я
p
W
3) ь, = 1
5) b, = 6, q = 2,
? = -j, и = 6%
•.‘.•.‘.'.'.V.’A’.VAV.AS
WA'AW.VAW.VA
'.S’A’AWAWAW
SS’AWAV.V.'A*
7) ^ = -8, <7 “ ; В) Ъх =12,
Л7 = 8 ;
/1 = 7;
« = 5 bo’lsa uning
dastlabki n
wv.v.v.v.
1) Ji, 1Щ
■.•AVAV.V.V.V
.•AV.SSW.V
WAVAW
.ir
1
2) 2, 6, 18, ... ; 3) 5, 1, 7,
•VA*.
’.VA'.
•X'.VAV . . ., .VAVi'A'A’.VA'A * Ч’Л'.уЛ'.у.'Л'/Л'
-v; ^ 8 — 3 2
/ V-WAWAW.VA' * * SW.VA /
-
V-SNV.
; 5)15, 5,
5
3
1) q = 2, S1 = 635 bo’lsa, bx va b1 ni toping.
2) q = -2 , 5g = 85 bo’lsa, bx va bs ni toping.
4. Agar geometrik progressiyada:
1)5,,=189, bx = 3, q = 2 ; 2) 5„ = 635 , 6, = 5 ,
3) 5,,=170, 6, =256, = 4)5„=99, bx =
q = 2 ■
-9, q = -2
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5) Sn = 441, bx = 7, q = 2 bo’Isa uning hadlari soni n ni toping.
5. Agar geometrik progressiyada:
1) Л = 7, 4 = 3,
2) 6, = 8, ? = 2,
3) *,=2, = 1458
4) ^=1, ^ = 2401
= 187 bo’lsa n va toping.
= 4088 bo’lsa n va toping.
S„ =2186 bo’lsa n va <7 toping.
S„ =2801 bo’lsa n va g toping.
6. Agar sonlar yig’indisining qo’shiluvchilari geometrik progressiyani
л<*. v!«.
ketma-ket hadlari bo’lsa, shu yig’indini toping.
1) 1 + 2 + 4 +... +128 ■ 2) 1 + 3 + 9 +... + 729 ■
3) -1 + 2-4 + ... + 128 ; 4) 5-15 + 45-... + 405 ;
5) 2 + 6 + 18 + ... + 1458 ; 6) 5 + 25 + 125 + ...+^5625 .
/MV.V.V.VANV.WW, .
II
w-s
«“■I
%
•WftW*
* * * CWteKWpw,
ян
7. Agar geometrik progressiyada:
1) 4, = 15, b3 = 25 ; 2)4,-14, 4, =
3) b2 = 3 , bs = 24 ; 4) 4,--5, 4Л-75 bo’Isa 4,, 5,ni toping;
8. Geometrik progressiyaning n - hadi formulasi bilan berilgan.
1) bn = 3-2”_1 bo’lsa, S5 ni
2) bn = 5 • 3"+2
v/Xvi/XvIv..
'ТОЙШ
‘.•AV.V>.W
/ 1 \
3)
b =-2-
«
2
Ж
. ,'lsa. Ц Ш toping.
4ni
V*NV«SV-%V«V.y.
Ш'
::«v
vav.';:**
Ж
f”
Ik bo’lllll^ ni toping.
1) W bo’lsa, 6, va q ni toping.
Xv.ft, vvXgSSXjXvvv NssswSss
.’XvXvX
2) — 12 В >S3 = 372 bo’lsa, q va £3 ni toping.
10. Agar geometrik progressiyada:
1) %У1*з + b5 = 90 bo’ Isa, ni toping.
2) b2 = §, b4+b6 = 60 bo’lsa, q ni toping.
3) bx — 63 =15 va b2 -b4 =30 bo’lsa, Sw ni toping.
4) b3 -bx = 24 va b5-bx = 624 bo’lsa, S5 ni toping.
11. Agar geometrik progressiyada:
l)*i=J, q = -4,n = 5 ; 2) 4, = 2, <7 = -j,n = 10 ;
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3)^1 =10, q = 3,n = 6 ; 4) bx = 5, q = -l, n = 9 ;
1
5) 6, =25, g = 1, n = & ; 6) ^=12, q--,n = 5 bo’lsa
uning dastlabki n ta hadining yig’indisini toping.
12. Geometrik progressiyaning dastlabki n ta hadining yig’indisini
toping.
1) 128, 64, 32,
n = 8 ; 2) 162, 54, 18,
n = 6
3>3
5)2,
2
10,
3
8’
50,
n = 5
n - 6
3 J_
4’ 2
6)20, 5,
4)
' 3’
1,25
л • • • •
n =1Л
.•X'XWSS
w
13. Agar geometrik progressiyada:
^7=16; 2) = -3, 66 = — 8||
1)
4)
b,=-
боб
=-I 56 =——
4 5’ 125
=m
= 1
• W.W.VSWAW.\W
V.SW.V.V.V.N'.*.'
ivsvs'
5) b: = 1| fh% = 96 ; ЩД = 1 ,bA = -21
bo’lsa uning dastlabki sakkizta hadlar|!yig’indisini toping.
: ga, dastlabki oltita
vwNweSwX
w.v.v.v.v.v .v
14. Dastlabki beshta hadining yig’indisi
hadining yig’indisi -126 g^Aima^aji|2 ga teng geometrik progres-siyaning
birinchi hadini toping.
15. Geometrik progreslllltiiig maxraji 3 ga, dastlabki to’rtta hadlari
yig’indisi 80 ga teng.: Uning to’rtinchi hadini toping.
1
W
W V.ViV.V.VA
.WWV.S*.
16. Geometrilgprogressiyaning birinchi hadi 486 ga, maxraji - ga
teng. Shu progressiyan||^dastlabki to’rtta hadi yig’indisini toping.
17. Geometrik prd|i|ssiyaning dastlabki uchta hadi yig’indisi -26
ga, dast||bki to’rttasiniki osa -80 ga teng. Agar shu progressiyaning
birinchi hadi - 2 ga teng bo’lsa, uning maxrajini qanchaga teng
mmm.
18. Gdbmetrik progressiyaning maxraji 3 ga, dastlabki to’rtta hadi¬
ning yig’indisi 80 ga teng. Birinchi hadining qiymatini toping.
19. Maxraji 2 ga teng bo’lgan geometrik progressiyaning dastlabki
oltita hadi yig’indisi 126 ga, dastlabki beshta hadi yig’indisi 62 ga
teng. Progressiyaning birinchi hadini toping.
20. Geometrik progressiyaning maxraji - 2 ga, dastlabki beshta hadi
yig’indisi 5,5 ga teng. Progressiyaning beshinchi hadini toping.
21. Agar olti hadli geometrik progressiyaning dastlabki uchta hadi¬
ning yig’indisi 112 ga va oxiridagi uchta hadining yig’indisi 14 ga
SOLIYEV X. X.
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«Ik. fill
•А'.У.'.'Л'Л'Л’ЛЧЧ'Л'ЛЧЧ'Л'Л*.*.-. v/.v,
teng bo’Isa, birinchi hadi nechaga teng bo’ladi?
22. Ishorasi almashinuvchi geometrik progressiyaning birinchi hadi
2 ga, uchinchi hadi 8 ga teng. Shu progressiyaning dastlabki 6 ta hadi
yig’indisini toping.
23. Geometrik progressiyaning barcha hadlarining yig’indisi uning
toq nomerli hadlari yig’indisidan 3 marta ko’p. Agar geometrik prog-
ressiya hadlarining soni juft bo’lsa, uning maxrajini toping.
24. (*„) geometrik progressiyada q = 2 va S4 = 5. b2 ni toping.
25. an arifmetik progressiyaning hadlari ayirmasi 1 ga teng.
\4\4v!|!v!viv!v
(ai ~ai)2 +(a5 -a3Y +- + (ai9~anY yig’indini hisoblang."^
%v ^.y.V.‘
26. O’suvchi geometrik progressiyaning dastlabki to’rtta hadi yig’in-
disi 15 ga, undan keyinga to’rttasiniki as a 240 ga teng. Shu progres¬
siyaning dastlabki oltita hadi yig’indisini topingiil:. |||
.: * ‘v!;X\v
27. 5 ta haddan iborat geometrik progressiyaning hadlari yig’indisi
birinchi hadini hisobga olmaganda 30 g% oxirgisini fiibbga olmaganda
15 ga teng. Shu progressiyaning ucbjlkbi hadini toping.
28. Geometrik progressiyada S6 ~:S5 '±n\2$ va q = -2. 68 ning
qiymatini toping.
29. Agar geometrik progressiyada Sk -Sk_{ = 64 va Sk+] - Sk = 128
bo’lsa, uning maxrajini qanchaga teng bo’ladi?
. Л * Vv.yA4W.yX%4V<\4W.J, XyX
30. Agar geometrik progressiyada Sk ~ SA._, = 28 va S
bo’lsa, uning maxrajini qanchaga teng bo’ladi?
■
31. Tenglamani yeching. 1 - x + x2 - x3 +... + xz - x9 =0
\V.\\\*.4W.4\\\\
*+i “ - 84
32. Tenglamani yeching#1 -3x + 9x2 -...-39 x9 = 0
sy.y.wy.v.v.v. ч*.*.Ч*
33. Arifmetik progr|§Siyaning hadlari au a2, a3f... s an ayirmasi
.. («2 -ar)+fc +(«4 -Й3)2 +-+(o„,i-a,,)"
34; Glti haddan iborat geometrik progressiyaning dastlabki uchta hadi-
- »•'%***
ning yig’indisi 168 ga, keyingi uchtasiniki 21 ga teng. Shu progressiya-
ning maxrajini toping.
35. Geometrik progressiya uchun quyidagi formulalardan qaysilari
to’g’ri?
»,(i-g")
\)b„=brq
2) К = 4_ • b
/1+1 5
3)
1 -q
36. Yig’indisi 15 ga teng bo’lgan uchta son arifmetik progressiya¬
ning dastlabki uchta hadidir. Agar bu sonlarga mos ravishda 1 ; 3 va
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9 sonlari qo’shilsa, hosil bo’lgan sonlar o’suvchi geometrik progres-
siyaning ketma-ket hadlari bo’ladi. Geometrik progressiyaning dast-
labki 6 ta hadi yig’indisini toping.
37. Agar hadlari haqiqiy sondan iborat bo’lgan o’suvchi geometrik
progressiyaning birinchi uchta yig’indisi 7, ko’paytmasi 8 ga teng
bo’Isa, shu progressiyaning beshinchi hadini toping.
38. Geometrik progressiyada 6, =-0,5 va <7 = 2 bo’lsa, Sl4 -Su
ayirmani toping.
39. Geometrik progressiyada bx = 1 va q = V2 bo’lsa,
bx + b3 + b5 +... + bl5 ning qiymatini hisoblang.
40. Geometrik progressiyada b{ = 2 va q = л/з bo’Isa,
b5 + b1 + b9 +... + b25 ning qiymatini
41. Geometrik progressiyaning dastlabki oltfta hadlyig’indisi 1820
ga, maxraji osa 3 ga teng. Shu progressiyaning birinchi va beshinchi
hadlari yig’indisini toping.
42. -0,25 ; 0,5 ;... geometrik progre^iyaping hadlari 10 ta. Shu
progressiyaning oxirgi 7 ta hadi yig
43. Geometrik progressiyaning ikkiiicbi hadi 2 ga, beshinchi hadi 16
ga teng. Shu progressiyaning dastlabki oltita hadi yig’indisini toping.
4i
ketma-ketlikning dastlabki n ta hadining
44.1;3;7;15;31-^Kkl;..1
yig’indisini topingJ||;
45. 2 ; 8 ; 26 ; 80 -Щ ketma-ketlikning dastlabki n ta
44.
A* V.V.VSWV.V.VA
•.'.S’A’.S'.V.V.VA
•.%V.V.V.V.V.\S4
V.VAWAWASN
‘.W.WV.SWA'A
* .’AW.V.WAvA
‘ЛЧ'ЛЧ,А*Л\,Л'А
\wKiwa\\
•Л-А-.ГЛХЧ'/.Л
VAWAWAW
.•AWAW.V
\VVAiA*A*
1 , 7
bx = 2 ; b„ = - va S„ = 3— bo’lsa,
иеотетк progressiyaning oltinchi va birinchi hadi ayirmasi
(|j|i|jmaxraji 3 ga teng. Shu progressiyaning dastlabki beshta hadi
yig indisini toping.
46. Geometrik progressiyaning birinchi hadi va maxraji 2 ga teng. Shu
progressiyaning dastlabki nechta hadlari yig’indisi 1022 ga teng bo’ladi?
4. Cheksiz kainayuvclii geometrik progression,
1. Geometrik progressiya cheksiz kamayuvchi bo’lishini isbotlang.
11 111
1)1,
2 ’ 4 ’
2)
3 , 9, 27 ’
3) -81, -27, -9, ... ;
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]_
5’
4) -16, -8, -4, ... ; 5)16, 8, 4, ... ; 6)5, 1,
2. Agar geometrik progressiyada:
1)6, =40, b2 = -20; 2) 67=12, *„=7; 3) 67 =-30 , 66=15;
4) b5=-9,
b9=-
1
27
;5)63 = 8, 66=2 ; 6) 68 = -15 ,
^10 “ 9
bo’lsa, u cheksiz kamayuvchi bo’ladimi. Shu aniqlang?
3. Cheksiz kamayuvchi geometrik progressiya yig4ndis|h||bpii)g.
...
v.v.v.v
да
1) 1,
]_
3’
l
9’
4)-7, -1,
7 ’
2) 6, 1, .
•• ; 5) 2, j
V'.N4V>SS*,-1V.S,,V.S,.4*.V.V.VA,.>
' N\4\y.v.sv.w.\\y.vAw.sv«.v
m'AW.JAW.W.VAWA’A
ШШ:
3) “25 -5, -1
1
50
,АГ W
«11
• •.v.v.v.v.v.vawawa*. •
• 4%V.%V.4*.V.V,V.S4V.*
* NWAWA'iW.*
« •VA'AV.VI.NSV
Л. N WAWSV
W svvw.»
'*>• • «»■
,w V.v
V.V.V.*
r
•:«v.
.•.v.v.v.y.
.VAWAW. .
.'.•,4V.\VV.NW.
4. Agar cheksiz kamayuvchi geometrik progress iyadi|||P
. 1 1 m. , %j.t 1
*•=- <? = -: 2)^=9, ; 3)*,=-. <7 = t
. illilihsA
1)
4)
1
8
b*~l
1
q
2’
-S'.V.VAV.'.V.N
* • ■ •
81
5) 6^;||||12
УЩ1| , 1
q = о, ——
1 1 С >
1
ШШЬоВи
bo’lsa, uning hadlari yig’i
5. n - hadining formulas! bilan
kamayuvchi geomqt$k prbgr||||ya tfo’la oladimi?
AvXvi^'lv.X‘A'. 4vXwX'!v///Xv!
Xsnsnsssssssss:*v«v VAV.V.V.V.gW/
lir /
2
9=-?
AV.4V
v.v.v.-.-*\v.v. Д J T.'.V.*
v.w\*.«ev\v.\v Wv.'.V
sv.;.v.\;av.;.v. ,.;л;л
quyidagi ketma-ketlik cheksiz
SV
1 )bn
4)
b =
/1
/2+1
‘.VA'.V.'.’.SV.’.V.*
3)
K = 2
•PI^.
.•.S'.W'.'.V.V.V.
4W.V.V.V.V,V.
w.4.vav.S|-.S’.
_ J.VV.-VA’/11 •‘i.
. WA’.V.V.V.'
w
1
llsf*,.=T'(-5)"
V
2
3
\
У
/2-1
6)
A =8
.Ч'Л'ЛЧ^Л'.
'V
v7>
/2+3
1)
Cheksiz kamayuvchi geometrik progressiyada:
V.V.V.
4)
65 =
16
... #*'•'
32
243
1
q = — ■
2 ’
2
q = — •
3 ’
2)
? =
3)
6. =
1
48
q
4 ’
5)ЬЪ =
2V2
27
? =
bo’lsa, yig’indisini toping.
7. Cheksiz kamayuvchi geometrik progressiyaning yig’indisi 150 ga teng,
Agar:
1
1) 4 - — bo’lsa, 6, ni; 2) 6, = 75 bo’lsa <7 ni toping.
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8. Berilgan geometrik progressiya cheksiz kamayuvchi ekanligini
isbotlang va uning yig’indisini toping.
1 1 1 1 1
1)-
3)1,
2 5
1
8’
2)-l,
16 ’
4 ’
_ I -i i __L
3 ’ 9 ’ 3 ’ 65 12 ’
9. Agar geometrik progressiyada:
1)*2=-81, S2 =162 ; 2)6, =33, S2 = 67 ;
3) 6, + b3 = 130 , 6, -b3 = 120 ; 4) b2 +b4 = 68 , b2 - ^ 60 t»o’
u cheksiz kamayuvchi ekanligini ko’rsating.
A
AVAWA
iVAWA*.
AV.WA
.•.S’AV.'.
.V.SW.'.
.•AWA*.
... v.v.v
r :•*"
10. Geometrik progressiyaning yig’indisini toping. V5 ; 1 ; ; ...
11. Cheksiz kamayuvchi geometrik progressiyaning hadlari yig’indisi
8 ga, dastlabki to’rtasiniki asa 7,5 ga teng. Agar uning barcha hadlari musbat
bo’lsa, progressiyaning birinchi hadipi'toping.
12. Cheksiz kamayuvchi geometrik progressiyaning hadlari yig’indisi
12 ga, maxraji esa - 0,5 ga teng. Uning birinchi va ikkinchi hadlari
ayirmasini toping.
13. Cheksiz kamayuvchi gepmetrili;progressiyaning yig’indisi 9 ga,
'iV/.V/.y
maxraji esa ~ ga teng. ВЩЩЬдппсШ hamda uchinchi hadlarining ayirmasini
toping.
14. Cheksiz kamayuvchi geometrik progressiyaning birinchi hadi
ikkinchisidan 8 ga ortiq, j^larining yig’indisi esa 18 ga teng.
Progressiyaning uchinchi hadini toping.
15. Cheksiz kamayuvchi geometrik progressiyaning birinchi hadi 3
gafhadlarining yig’indisi osa 4,5 ga teng. Shu progressiyaning
•.v.x.v.v.:.v.Xv\..
16/Gheksiz kamayuvchi geometrik progressiyaning yig’indisi 56 ga, hadlari
kvadratlari yig’indisi 448 ga teng. Progressiyaning maxrajini
toping.
17. Cheksiz kamayuvchi geometrik progressiyaning yig’indisini
1
toping.
11111
— 1 1
2 3 4 9 8 27
+ ...
18. Hisoblang.
1
h ...
81
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17*
+ - cheksiz kama-
19. Tenglamani yeching. (0 < * < 2)
*3 (l + (l - *)+ (l - *)2 + (l - *)3 + ...) = 2221-1
20. a ning qanday qiymatida 2я + a42 + a + -j=
yuchi geometrik progressiyaning yig’indisi 8 ga teng bo’ladi?
21. Cheksiz kamayuvchi geometrik progressiyaning hadlarining yig’in¬
disi uning dastlabki ikkita hadi yig’indisidan 2 ga ko’p. Progressiyaning
birinchi hadi 4 ga teng. Shu progressiyaning hadlari yig’indisini topingl
22. Bir-biridan faqat maxrajlarining ishoralari bilan farq qiiadigari;2:ta
cheksiz kamayuvchi geometrik progressiya berilgan. Ulaming yig’in-
dilari mos ravishda S{ va S2 ga teng. Shu progressiyalardan istalgani-
ning hadlari kvadratlaridan tuzilgan cheksiz kamayuvchi geometrik
progressiyaning yig’indisini toping.
23. Cheksiz kamayuvchi geometrik progressiyanirig yig’indisi 243 ga,
dastlabki beshta hadiniki osa 275 gateng. Bu progressiyaning maxraji
0,125 dan qanchaga kam?
• SW.V.V.V.V.V.V.VA .
24. Hisoblang.
25. Hisoblang.
26. Hisoblang^
ШЁ/Г
pKR»
•.‘.•Л'Л'Л'.Ч'Л*.'
'.•.•.•.•.'.•.•.•.'.•.•Л
* V.'.S'.'.'X'.*
■■•Ч-'Г
27 • m+w+-
ъщш-ш
28. Hadlarining yig’indisi 2,25 ga, ikkinchi hadi 0,5 ga teng bo’Isa,
cheksiz kamayuvchi geometrik progressiyaning maxraj ini toping.
2fe$:CMksil-ili!lavuvchi geometrik progressiyaning birinchi hadi 2 ga,
haiiHiii||yig’indisi osa 5 ga teng. Shu progressiyaning hadlari
kvadratlaridan tuzilgan progressiyaning hadlari yig’indisini toping.
30. Tenglamani yeching. (j^r| < l), — +1 + * + x2 + ... + *”+... = 4,5
31. Cheksiz kamayuvchi geometrik progressiyaning hadlari yig’indisi
1,6 ga, 2-hadi - 0,5 ga teng. Shu progressiyaning uchinchi hadini toping.
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