|
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
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