109 Ko‘paytuvchilarga ajrating
(355 — 360) :
355. 1)
(
);
a b c a b + +
+
3)
3 (
)
;
x a x y y +
+
+
2)
(
);
m n p m n - +
-
4)
2 (
)
.
x a x y y +
-
-
356. 1)
(
)
2;
(
)
x y x y +
+
+
3)
(
) (
)
-
+
-
2
2
;
m m n m n 2)
(
)
2
;
a b a b -
+ -
4)
(
) (
)
- +
-
2
4
1
1 .
q p p 357. 1)
(
)
2
;
m m n m n -
+ -
3)
2 (
)
;
m m n n m -
- +
2)
(
)
4
1
1;
q p p - + -
4)
(
)
4
1 1
.
q p p - + -
358. 1)
(
)
;
a x c bc bx - +
-
3)
(
)
3 2
8
4 ;
a b c b c +
+
+
2)
(
)
;
a b c db dc +
+
+
4)
359. 1)
2
2 ;
ac bc ad bd +
-
-
3)
2
3
6
;
bx ay by ax -
-
+
2)
3
3 ;
ac bd ad bc -
+
-
4)
5
3
15 .
ay bx ax by -
+
-
360. 1)
-
-
+
+
-
2
2
2
;
xy by ax ab y a 2)
-
-
+
+
-
2
2
2
.
ax ay bx cy by cx 361. Hisoblang:
1)
139 15 18 139 15 261 18 261;
×
+
×
+
×
+
×
2)
125 48 31 82 31 43 125 83;
×
-
×
-
×
+
×
3)
14,7 13 2 14,7 13 5,3 2 5,3;
×
- ×
+
×
- ×
4)
×
+
× +
×
+
×
1
1
2
1
4
2
3
5
3
3
5
3
3
4
4,2
3
2
2,8
.
362. Ifodaning son qiymatini toping:
1)
-
-
+
= -
=
2
5
5
7
7 , bunda
3,
4;
a ax a x x a 2)
-
-
+
=
=
2
3
3 , bunda
0,5,
0,25;
m mn m n m n 3)
+
-
-
=
=
2
5
5 , bunda
6,6,
0,4;
a ab a b a b 4)
-
-
+
=
=
a ab a b a b 2
7
20
2
2 , bunda
,
0,15.
M a s h q l a r
110 21- 363. Hisoblang:
1)
-
×
+
×
2
287
287 48 239 713;
2)
+
×
-
×
2
73,4
73,4 17,2 90,6 63,4.
364. Tenglamani yeching:
1)
(
)
4
4 0;
x x x -
+ - =
2)
(
)
7
4
28 0.
t t t +
-
-
=
Ali bilan Valining massasi birgalikda 5 ta tarvuz massasiga teng. Valining massasi 1 ta qovun massasidan 4 marta ko‘p. Vali bilan 2 ta qovunning birgalikdagi massasi 3 ta tarvuz massasiga teng. Alining massasi nechta qovunning massasiga teng? Yig‘indining kvadrati. Ayirmaning kvadrati Ikkita son yig‘indisining kvadrati (
a +
b )
2
ni qaraymiz. Ko‘p-
hadni ko‘phadga ko‘paytirish qoidasidan foydalanib, hosil
qilamiz:
(
) (
) (
)
2
2
2
2
2
2
,
a b a b a b a ab ab b a ab b +
=
+
+
=
+
+
+
=
+
+
ya’ni
(
)
2
2
2
2
.
a b a ab b +
=
+
+
(1)
Ikki son yig‘indisining kvadrati — birinchi son kvadrati, qo‘shuv birinchi son bilan ikkinchi son ko‘paytmasining ikkilangani, qo‘shuv ikkinchi son kvadratiga teng. (1) formulani 20- rasmda tasvirlangan kvadratning yuzini
ko‘zdan kechirib, osongina hosil qilish mumkinligini aytib
o‘tamiz.
Endi ikki son ayirmasining kvadratini qaraymiz:
(
) (
)(
)
-
=
-
-
=
-
-
+
=
-
+
2
2
2
2
2
2
,
a b a b a b a ab ab b a ab b ya’ni
(
)
2
2
2
2
.
a b a ab b -
=
-
+
(2)
¹ 7
111 20- rasm. a b a 2
ab ab b 2
b a Ikki son ayirmasining kvadrati — birinchi son kvadrati, ayiruv birinchi son bilan ikkinchi son ko‘paytma- sining ikkilangani, qo‘shuv ikkinchi son kvadratiga teng. (1) va (2) tengliklarda
a va
b istalgan
sonlar yoki algebraik ifodalardir.
(1) va (2) formulalarni qo‘llashga
doir misollar:
1)
(
) ( )
( )
+
=
+ ×
×
+
=
+
+
2
2
2
2
2
2
3
2
2 2
3
3
4
12
9 ;
m k m m k k m mk k 2)
(
) ( )
-
=
- ×
× +
=
-
+
2
2
2
2
2
2
4
2
5
3
5
2 5
3 3
25
30
9;
a a a a a 3)
(
)
( ) (
)
(
)
( ) (
)
(
)
( )
2
2
2
2
2
2
2
2
2
3
1
3
1
3
3
2 3
3
6
9 .
a b a b a b a b a a b b a ab b - -
= -
+
= -
+
=
=
+
=
+
×
+
=
+
+
Zaruriy hisoblashlarni og‘zaki bajarib, oraliq natijalarni
yozmaslik mumkin. Masalan, birdaniga bunday yozish mum-
kin:
(
)
2
2
2
4
2 2
4
5
7
25
70
49 .
a b a a b b -
=
-
+
Yig‘indi yoki ayirmaning kvadrati formulalari
qisqa ko‘pay- tirish formulalari deyiladi va ba’zi hollarda hisoblashlarni sod-
dalashtirish uchun qo‘llaniladi. Masalan:
1)
(
)
=
-
=
-
+ =
2
2
99
100 1
10 000 200 1 9801;
2)
(
)
=
+
=
+
+ =
2
2
52
50 2
2500 200 4 2704.
(1) formula (1 +
a )
2
ifodaning qiymatlarini taqribiy hi-
soblashlarda ham qo‘llaniladi.
a son musbat yoki manfiy son
bo‘lib, uning moduli 1 ga nisbatan kichik bo‘lsa (masalan,
a = 0,0032 yoki
a =
-
0,0021), u holda
a 2
son yanada kichik
bo‘ladi va shu sababli
(1 +
a )
2
= 1 + 2
à +
à 2
tenglikni (1+
a )
2
»
1+2
a taqribiy tenglik bilan almashtirish
mumkin. Masalan:
112 M a s h q l a r 1
) (1,002)
2
=
(1 + 0,002)
2
»
1 + 2 · 0,002
=
1,004;
2) (0,997)
2
=
(1
-
0,003)
2
»
1
-
2 · 0,003
=
0,994.
Yig‘indining kvadrati va ayirmaning kvadrati formulalari
ko‘phadni ko‘paytuvchilarga ajratishda ham qo‘llaniladi, ma-
salan:
1)
(
)
+
+
=
+ × × +
=
+
2
2
2
2
10
25
2 5
5
5 ;
x x x x x 2)
( )
( ) (
)
-
+
=
- ×
×
+
=
-
2
2
2
4
2 3
6
2
2
3
3
2
3
8
16
2
4
4
4
.
a a b b a a b b a b Masala. Formulani isbotlang:
(
)
3
3
2
2
3
3
3
.
a b a a b ab b +
=
+
+
+
(3)
(
) (
) (
) (
)
(
)
+
=
+
+
=
+
+
+
=
3
2
2
2
2
a b a b a b a b a ab b =
+
+
+
+
+
=
+
+
+
3
2
2
2
2
3
3
2
2
3
2
2
3
3
.
a a b ab a b ab b a a b ab b Xuddi shunga o‘xshash,
(
)
3
3
2
2
3
3
3
a b a a b ab b -
=
-
+
-
(4)
formulani ham isbotlash mumkin.
(3) va (4) formulalar, mos ravishda,
yig‘indining kubi va ayirmaning kubi formulalari deb ataladi.
(3) va (4) formulalar ham
qisqa ko‘paytirish formula- lari hisoblanadi.
Quyidagi mashqlarda ikkihadning kvadratini ko‘phad shak-
lida tasvirlang
(365 — 372) :
365. 1)
(
)
+
2
;
c d 3)
(
)
+
2
2
;
x 5)
(
)
+
2
3 ;
y 2)
(
)
-
2
;
x y 4)
(
)
+
2
1 ;
x 6)
(
)
+
2
7
.
m 366. 1)
(
)
-
2
2 ;
m 3)
(
)
-
2
7
;
m 5)
+
2
1
3
;
a 2)
(
)
-
2
3 ;
x 4)
(
)
-
2
6 ;
y 6)
+
2
1
2
.
b
113 367. 1)
(
)
+
2
2
;
q p 2)
(
)
+
2
3
2
;
x y 3)
(
)
-
2
6
4
;
a b 4)
(
)
2
5
.
z t -
368. 1)
(
)
+
2
2
3
1 ;
a 2)
(
)
+
2
2
1 ;
a 3)
(
)
+
2
2
2
2
3
;
x n 4)
(
)
+
2
2
2
.
x y 369. 1)
-
2
1
5
;
m 2)
-
2
1
3
;
a 3)
-
2
2
3
;
a b 4)
+
æ
ö
ç
÷
è
ø
2
3
4
.
y x 370. 1)
(
)
+
2
0,2
0,3
;
x y 3)
-
2
3
2
3
3
4
;
x 2)
(
)
-
2
0,4
0,5
;
b c 4)
-
2
3
1
4
4
5
.
a 371. (
)
+
=
+
+
+
3
3
2
2
3
3
3
a b a a b ab b formulaga qanday geometrik
ma’no bera olasiz?
Nuqtalar o‘rniga mos so‘zlarni qo‘ying:
Qirrasining uzunligi
a va
b bo‘lgan ... yasaymiz. O‘l-
chamlari
a ½
a ½
b va
a ½
b ½
b bo‘lgan .... yasaymiz. Ularni
shunday taxlaymizki, ... hosil bo‘ladi.
372. 1)
(
)
-
-
2
2
4
5
;
ab a 3)
(
)
+
2
2
0,2
5
;
x xy 2)
(
)
-
-
2
2
3
2
;
b ab 4)
(
)
+
2
2
4
0,5
.
xy y Qisqa ko‘paytirish formulalaridan foydalanib, amallarni
bajaring