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
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)
Do'stlaringiz bilan baham: