|
13.
7 ga karrali dastlabki nechta natural sonni qo‘shganda 385 hosil
bo‘ladi?
A) 12; B) 11; C) 10; D) 55; E) 56.
14.
Geometrik progressiyada
b
1
= 2,
q
= 3 bo‘lsa,
S
6
ni toping.
A) 1458; B) 729; C) 364; D) 728; E) to‘g‘ri javob berilmagan.
15.
Geometrik progressiyada
3
1
=
q
,
S
= 364 bo‘lsa,
b
1
ni toping.
A)
3
2
63 ; B) 81; C)
3
1
121 ; D) 240; E)
2
3
242 .
16.
Geometrik progressiyada
8
5
4
10
=
S
,
8
5
5
42
=
S
,
b
=
1
1
8
bo‘lsa,
q
ni
toping.
A) 4; B) 2; C) 8; D)
2
1
; E) 2 .
186
17.
Geometrik progressiyada 6 ta had bor. Dastlabki 3 ta hadining
yig‘indisi 26 ga, keyingi 3 tahadining yig‘indisi esa 702 ga teng.
Progressiya maxrajini toping.
A) 4; B) 3; C)
3
1
; D)
3
2
; E)
3
4
.
18.
Cheksiz kamayuvchi geometrik progressiyada
4
1
1
=
b
,
S
= 16
bo‘lsa,
q
ni toping.
A)
2
1
; B)
65
64
; C)
64
63
; D)
4
1
; E)
8
1
.
19.
Geometrik progressiyada
2
3
=
q
,
b
1
=
3
2
-
bo‘lsa,
S
ni toping.
A)
3
2
+
; B) 3; C)
3
3
2
; D) 2; E) 3 .
1
.
Beruniy masalasi.
Agar hadlari musbat geometrik pro-
gressiyaning: hadlari soni toq bo‘lsa, u holda
k
k
b
b b
+
+
=
×
2
1
1
2
1
;
hadlari soni juft bo‘lsa,
k
k
k
b b
b b
+
×
=
×
1
1
2
bo‘lishini isbotlang.
2
.
Axmes papirusidan olingan masala (eramizdan oldingi
2000- yillar).
10 o‘lchov g‘allani 10 kishi orasida shunday
taqsimlaginki, bu kishilarning biri bilan undan keyingisi
(yoki oldingisi) olgan g‘alla farqi
1
8
o‘lchovga teng bo‘lsin.
«Qadimgi xalqlardan qolgan yodgorliklar» asarida Abu
Rayhon Beruniy shaxmatning kashf etilishi haqidagi rivoyat
bilan bog‘liq birinchi hadi
b
1
=
1 va maxraji
q
=
2 bo‘lgan
geometrik progressiyaning birinchi 64 ta hadining yig‘indi-
sini hisoblaydi; shaxmat taxtasidagi
k
- katakka mos sondan
1 soni ayirilsa, ayirma
k
- katakdan oldingi barcha kataklarga
mos sonlar yig‘indisiga teng bo‘lishini ko‘rsatadi, ya’ni
q
k
-
1
=
1
+
q
+
q
2
+
...
+
q
k
-
1
ekanini isbotlaydi.
&
T a r i x i y m a ’ l u m o t l a r
6
T a r i x i y m a s a l a l a r
187
IX SINF «ALGEBRA» KURSINI TAKRORLASH
UCHUN MASHQLAR
491.
Funksiyaning grafigini yasang:
1)
=
+
+
2
6
9;
y
x
x
2)
=
-
2
7
2
;
y
x
3)
=
-
+
2
12
4;
y
x
x
4)
=
+
-
2
3
1;
y
x
x
5)
=
+
2
;
y
x
x
6)
=
-
2
;
y
x
x
7)
=
-
+
(
2)(
5);
y
x
x
8)
æ
ö
=
+
+
ç
÷
è
ø
1
8
(
4).
y
x
x
492.
(Og‘zaki.)
y = ax
2
+
bx + c
funksiya grafigidan foydalanib (82- rasm),
uning xossalarini aniqlang.
493.
Funksiyaning grafigini yasang va xossalarini aniqlang:
1)
= -
-
-
2
2
8
8;
y
x
x
2)
2
3
12
16;
y
x
x
=
+
+
3)
=
-
+
2
2
12
19;
y
x
x
4)
= +
-
2
3 2
;
y
x x
5)
= -
-
2
4
4 ;
y
x
x
6)
=
-
-
2
12
4
9.
y
x
x
494.
Funksiyaning grafigini bitta koordinata tekisligida yasang:
1)
=
= -
2
2
1
1
3
3
va
;
y
x
y
x
2)
=
=
-
2
2
3
va
3
2;
y
x
y
x
3)
= -
= -
+
2
2
1
1
2
2
va
(
3) ;
y
x
y
x
4)
=
=
-
+
2
2
2
va
2(
5)
3.
y
x
y
x
Tengsizlikni yeching (
495–499
):
495.
1)
-
+
>
(
5)(
3)
0;
x
x
2)
+
+
<
(
15)(
4)
0;
x
x
3)
-
+
£
(
7)(
11)
0;
x
x
4)
-
-
³
(
12)(
13)
0.
x
x
82- rasm.
188
496.
1)
x
2
+ 3
x
> 0;
2)
-
<
2
5
0;
x
x
;
3)
-
£
2
16
0;
x
4)
x
2
– 3 > 0.
497.
1)
-
+ >
2
8
7
0;
x
x
2)
x
2
+ 3
x
– 54 < 0;
3)
+
- >
2
1
2
0,5
1
0;
x
x
4)
+
- <
2
5
9,5
1
0;
x
x
5)
-
-
+ >
2
3
4
0;
x
x
6)
-
+
- £
2
8
17
2
0.
x
x
498.
1)
-
+ >
2
6
9
0;
x
x
2)
-
+
£
2
24
144
0;
x
x
3)
-
+ <
2
1
2
4
8
0;
x
x
4)
+
+
³
2
1
3
4
12
0;
x
x
5)
-
+ >
2
4
4
1
0;
x
x
6)
+
+ <
2
1
5
5
2
0.
x
x
499.
1)
2
10
30
0;
x
x
-
+
>
2)
-
+ - <
2
1
0;
x
x
3)
+
+ <
2
4
5
0;
x
x
4)
-
+
>
2
2
4
13
0;
x
x
5)
-
+ <
2
4
9
7
0;
x
x
6)
-
+
-
<
2
11 8
2
0.
x
x
Tengsizlikni intervallar usuli bilan yeching (
500–502
):
500
. 1)
+
-
>
(
3)(
4)
0;
x
x
2)
æ
ö
-
+
<
ç
÷
è
ø
1
2
(
0,7)
0;
x
x
3)
-
+
<
(
2,3)(
3,7)
0;
x
x
4)
+
-
£
(
2)(
1)
0;
x
x
501.
1)
+
-
³
(
2)(
1)
0;
x
x
2)
+
-
£
2
(
2)(
1)
0;
x
x
3)
+
-
>
2
(
2)(
1)
0;
x
x
4)
-
+
³
2
(2
)(
3 )
0.
x x
x
502.
1)
-
+
³
3
2
0;
x
x
2)
+
-
£
0,5
2
0;
x
x
3)
-
+
<
(
1)(
2)
0;
x
x
x
4)
+
-
<
2
(3
)(1
)
0.
x
x
x
503
. Trapetsiyaning yuzi 19,22 sm
2
dan ortiq. Uning o‘rta chizig‘i
balandligidan ikki marta katta. Trapetsiyaning o‘rta chizig‘ini
va balandligini toping.
504
. 320 m dan ortiq balandlikda uchib ketayotgan samolyotdan geolog-
larga yuk tashlab yuborildi. Yuk qancha vaqtda yerga kelib tusha-
di? Erkin tushish tezlanishi 10
m/s
2
ga teng deb qabul qiling.
189
505.
Parallelogrammning tomoni shu tomonga tushirilgan baland-
likdan 2 sm ortiq. Agar parallelogrammning yuzi 15 sm
2
dan
ortiq bo‘lsa, shu tomonning uzunligini toping.
506
. Tengsizlikni intervallar usuli bilan yeching:
1)
+
+
-
+
>
(
2)(
5)(
1)(
4)
0;
x
x
x
x
2)
+
+
-
+
<
2
(
1)(3
2)(
2)(
7)
0
x
x
x
x
;
3)
-
-
+
+
+
³
3
1
3
3
1
3
2;
x
x
x
x
4)
2
1 3
1 3
12
1 3
3
1
1 9
.
x
x
x
x
x
-
+
+
-
-
+
³
507.
Agar
x
2
+
px
+
q
kvadrat uchhad
x =
0 bo‘lganda –14 ga teng
qiymatni,
x = –
2
bo‘lganda esa –20 ga teng qiymatni qabul qilsa,
shu kvadrat uchhadning
p
va
q
koeffitsiyentlarini toping.
508
. Agar
y = x
2
+ px + q
parabola:
1) abssissalar o‘qini
= -
=
1
2
2
3
va
x
x
nuqtalarda kessa;
2) abssissalar o‘qi bilan
x
= –7 nuqtada urinsa;
3) abssissalar o‘qini
x
= 2
va ordinatalar o‘qini
y =
–1 nuqtada
kesib o‘tsa,
p
–
q
ni toping.
509.
Agar parabola abssissalar o‘qini 5 nuqtada kessa va uning uchi
æ
ö
ç
÷
è
ø
3
1
4
8
2 ; 10
nuqta bo‘lsa, shu parabolaning tenglamasini yozing.
510.
Teleskopning (reflektorning) qaytaruv-
chi ko‘zgusi o‘q kesimi bo‘yicha parabola
shakliga ega (83- rasm). Shu parabola-
ning tenglamasini yozing.
511.
Agar
y = ax
2
+
bx + c
kvadrat funksiya-
ning grafigi:
1)
A
(–1; 0),
B
(3; 0) va
C
(0; –6) nuqta-
lardan o‘tsa;
2)
K
(–2;0),
L
(1;0),
M
(0; 2) nuqtalardan
o‘tsa, uning koeffitsiyentlarini toping.
512.
Istalgan nomanfiy
a
va
b
sonlar uchun
1)
+
£
+
2
2
2
(
) ;
a
b
a b
2)
+
£
+
3
3
3
(
) ;
a
b
a b
3)
+
³
+
3
3
2
2
;
a
b
a b ab
4)
3
3
3
(
)
4(
)
a b
a
b
+
£
+
tengsizlikning to‘g‘ri bo‘lishini isbotlang.
83- rasm.
190
513.
Istalgan musbat
a
,
b
,
c
sonlar uchun
1)
3;
a
b
c
b
c
a
+ +
³
2)
;
a
b
c
bc
ac
ab
a b c
+
+
³ + +
3)
3
3
3
2
2
2
3
;
a
b
c
a b c
a b c
+ +
+ +
+ +
³
4)
3
2
a
b
c
b c
c a
a b
+
+
+
+
+
³
tengsizlikning to‘g‘ri ekanini isbotlang.
514.
Funksiyaning grafigini yasang:
1)
2
;
y
x
=
2)
|
1 |;
y
x
=
-
3)
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