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Ko‘phadni birhadga ko‘paytirish
Istalgan ko‘phadni birhadga ko‘paytirish ham xuddi shun-
day bajariladi, masalan:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
-
-
=
-
+ -
-
=
= -
+
-
+
-
=
-
-
-
+
+
-
= -
+
-
2
2
2
2
3
2
2
3
2
2
2
2
2
2
3
2
3
4
2
4
3
4
8
12
;
3
4
5
5
3
5
4
5
5
5
15
20
25
.
n m
nm
nm
n m
nm
nm
nm
n m
n m
a
ab
c
bc
a
bc
ab
bc
c
bc
a bc
ab c
bc
Ko‘phadni birhadga ko‘paytirish uchun ko‘phadning har
bir hadini shu birhadga ko‘paytirish va hosil bo‘lgan ko‘payt-
malarni qo‘shish kerak.
Ko‘phadni birhadga ko‘paytirish natijasida yana ko‘phad
hosil bo‘ladi. Hosil bo‘lgan ko‘phadni uning barcha hadlarini
standart shaklda yozib, soddalashtirish kerak. Oraliqdagi nati-
16-
12- rasm.
b
3a
a
2b
c
O‘lchamlari 12- rasmda ko‘rsatilgan
to‘g‘ri burchakli parallelepiðedni qaray-
miz. Uning hajmi asosining yuzi bilan
balandligining ko‘paytmasiga teng:
(
a
+ 2
b
+
c
)(3
ab
).
Bu ifoda
a
+ 2
b
+
c
ko‘phad bilan 3
ab
birhadning ko‘paytmasi bo‘ladi.
Ko‘paytirishning taqsimot qonunini
qo‘llab, bunday yozish mumkin:
(
a
+ 2
b
+
c
)(3
ab
) =
a
(3
ab
) + 2
b
(3
ab
) +
+
c
(3
ab
) = 3
a
2
b
+ 6
ab
2
+ 3
abc
.
85
jalarni yozmasdan, birhadlarni og‘zaki ko‘paytirib, birdaniga
javobni yozish ham mumkin, masalan,
(
)
2
2
2
2
3
3
1
3
2
2
3
2
4
2
.
æ
ö
-
+
-
-
=
-
+
ç
÷
è
ø
ab
ab
a
b
a b
a b
ab
Birhadni ko‘phadga ko‘paytirish ham shunga o‘xshash ba-
jariladi, chunki ko‘paytuvchilarning o‘rinlarini almashtirish
bilan ko‘paytma o‘zgarmaydi, masalan, 4
pq
(3
p
2
-
q
+ 2) =
= 12
p
3
q
-
4
pq
2
+ 8
pq
.
Ko‘phad va birhad ko‘paytmasini toping
(274
—
278):
274.
1)
( ) (
)
-
m
5 · 10 +
;
3)
(
)
-
y
1
7
2 -5 ·
;
2)
(
)
-
-
x
1 ·
2
2 +
;
4)
(
) (
)
-
-
m
n
2 +3 · 10 ;
275.
1)
(
)
;
-
a b n
3)
(
)
6 5
2 ;
-
-
x y
x
2)
(
)
5
2 ;
-
x + y
z
4
4)
(
)
2
1 .
- +
x
x
x
276.
1)
(
)
7
2
3 ;
+
ab a
b
3)
(
)
-
2
2
2
12
;
p q q p q
2)
(
)
+
2
5
15
3 ;
a b
b
4)
(
)
-
2
3
3
2
.
xy xy
x
277.
1)
(
)
17 5
6
;
+
-
a a
b
ab
3
3)
(
)
+
+
2
3
5
6
7 ;
x y x
y
z
2)
(
)
2
8
3
;
-
+
ab b
ac c
2
4)
(
)
+
+
2
2
2
2
3
.
xyz x
y
z
278.
1)
-
a b
ab
a b
3 2
4
3
1
3
4
2
4
3
;
2)
+
2 4
3
3
2
1
3
3
2
2
.
a b
a b
ab
M a s h q l a r
(
)
2
2
2
2
3
3
1
3
2
2
3
2
4
·
2
.
-
+
-
-
=
-
+
ab
ab
a
b
a b
a b
ab
86
Ifodani soddalashtiring
(279
—
281):
279.
1)
(
) (
)
2
3 3
2 ;
-
-
-
6
3
t
n
t
n
3)
(
) (
)
2 3
2
5 2
3
;
-
-
-
-
x
y
y
x
2)
(
)
(
)
4
;
-
-
-
5
2
3
a b
a
b
4)
(
)
(
)
6 5 7
.
-
+
7 4
3
p+
p
280.
1)
(
)
(
)
-
-
-
2
2
1 3
2 2 ;
x
x
x
x
2)
(
)
(
)
-
-
-
2
2
4
3 2
3
4 3 ;
a
b
b
a
b b
3)
(
) (
)
(
)
2 3
4
3
7
7 2
7 ;
a
+
+
-
-
-
a
a
4)
(
) (
)
(
)
5
3
6 3
4 .
- -
- +
-
3 2
1
x
x
x
281.
1)
(
)
(
)
(
)
5
0,
0,7
8 0,7 0,4
;
-
-
+
-
0,8
1
4
1
y
y +
y
2)
-
+
+
1
1
2
4
2
x
x
1
1
2
2
3
1
;
3)
-
-
-
x
x
5 1
1
4 1
3
4 5
5
5 4
4
;
4)
(
)
(
)
(
)
4
1,3
5 0,1
1,62 .
+
-
-
+
-
0,2 5
6
0,25
y
y
y
282.
Algebraik ifodaning qiymatini toping:
1)
(
)
(
)
+
-
+
=
= -
7 4
3
6 5
7 , bunda
2,
3;
a
b
a
b
a
b
2)
(
) (
)
+ -
-
-
2
1
2
1 , bunda =10, = 5;
a b
b a
a
b
3)
(
)
(
)
-
+
-
=
= -
2
2
2
2
3
4
4
3
, bunda
10,
5;
ab a
b
ab b
a
a
b
4)
(
)
(
)
-
-
+
= -
= -
2
2
4
5
3
5
4
, bunda
2,
3.
a
a
b
a
a b
a
b
Ko‘phadni ko‘phadga ko‘paytirish
Ushbu masalani qaraylik.
Masala.
O‘lchamlari 13- rasmda ko‘rsatilgan shkaflar bi-
lan to‘silgan devor sirtining yuzini toping.
Shkaflar bilan band bo‘lgan devorning sirti tomonlari
2
a
+
c
+ 2
a
= 4
a
+
c
va
a
+
b
+
a
= 2
a
+
b
bo‘lgan to‘g‘ri to‘rtburchakdan iborat. Bu to‘g‘ri to‘rtburchak-
ning yuzi
S
= (4
a
+
c
)(2
a
+
b
) ga teng.
(4
a
+
c
)(2
a
+
b
) ifoda (4
a
+
c
) va (2
a
+
b
) ko‘phadlarning
ko‘paytmasidir.
17-
87
Sonlarni ko‘paytirishning taqsimot qonunini qo‘llab,
S
= (4
a
+
c
)(2
a
+
b
)
= 4
a
(2
a
+
b
)
+
c
(2
a
+
b
)
kabi yozish mumkin. So‘ngra, 4
a
(2
a
+
b
) = 8
a
2
+ 4
ab
va
c
(2
a
+
b
) = 2
ac
+
bc
bo‘lgani uchun
S
= 8
a
2
+ 4
ab
+ 2
ac
+
bc
.
13- rasm.
Shunday qilib, mazkur ko‘phadlarning ko‘paytmasini to-
pish uchun 4
a
+
c
ko‘phadning har bir hadini 2
a
+
b
ko‘p-
hadning har bir hadiga ko‘paytirish va natijalarni qo‘shishga
to‘g‘ri keldi. Ixtiyoriy ikkita ko‘phadni ko‘paytirish ham xuddi
shunday bajariladi, masalan,
(
) (
)
(
) (
)
(
) (
)
-
-
=
×
+
× -
+ -
×
+
+ -
× -
=
-
-
+
=
-
+
2
2
2
2
7
2
3
5
(7 ) (3 ) (7 )
5
2
(3 )
2
5
21
35
6
10
21
41
10
.
n
m
n
m
n
n
n
m
m
n
m
m
n
nm
mn
m
n
nm
m
Ko‘phadni ko‘phadga ko‘paytirish uchun birinchi ko‘p-
hadning har bir hadini ikkinchi ko‘phadning har bir hadiga
ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Ko‘phadni ko‘phadga ko‘paytirish natijasida yana ko‘phad hosil bo‘-
ladi. Hosil qilingan ko‘phadni standart shaklda yozish kerak.
2a
c
2a
a
b
a
(
) (
)
-
-
=
-
-
+
2
2
7
2
3
5
21
35
6
10
.
n
m
n
m
n
nm
mn
m
88
Masalan,
(
)(
)
-
+
-
=
-
-
+
+
+
-
=
-
-
+
-
2
2
2
2
2
4
3
5
10
2
20
4
15
3
10
2
20
19
3 .
a
b
c
b c
ab
ac
b
bc
bc
c
ab
ac
b
bc
c
Bir nechta ko‘phadni ko‘paytirishni navbatma-navbat ba-
jarish kerak, masalan,
(
)(
)(
)
(
)
(
)
+
+
-
=
+
+
-
=
=
-
+
-
+
-
=
-
-
2
2
3
2
2
2
2
3
3
2
3
2
3
3
2
3
3
3
9
2
6
7
6 .
a b a
b a
b
a
ab
b
a
b
a
a b
a b
ab
ab
b
a
ab
b
Ko‘phadlarni ko‘paytiring
(283
—
291):
283.
1)
(
) (
)
;
+
+
a
a
2
3
3)
(
)
(
)
;
+
-
m
n
6
1
2)
(
) (
)
1
4 ;
-
+
z
z
4)
(
) (
)
4
5 .
+
+
b
c
284.
1)
(
) (
)
4
3 ;
-
-
c
d
3)
(
) (
)
;
x + y
x +
1
2)
(
) (
)
10
2 ;
-
- -
a
a
4)
(
) (
)
.
- +
- -
p q
q
1
285.
1)
(
) (
)
;
2
1
4
x +
x +
3)
(
) (
)
3
2 2
1 ;
-
-
m
m
2)
(
) (
)
2
5
;
-
a+
a
3
4
4)
(
) (
)
3
4
.
-
-
5
p
q
p q
286.
1)
+
-
a
b
a
b
1
1
2
2
3
3 ;
3)
-
+
1
1
3
3
a
b
a
b
2
2 ;
2)
(
) (
)
0,3
0,3 ;
-
+
m m
4)
(
) (
)
.
-
0,2
0,5
0,2
0,5
a+
x
a
x
287.
1)
(
) (
)
;
+
+
a
b
a b
2
2
3)
(
) (
)
2
2
2
2
a
b
a b
+
+
;
2)
(
) (
)
2
2
2
2
5
6
6
5
;
x
y
x
y
-
-
4)
(
)
(
)
2
2
1
3 .
x
x
x
+
+
+
288.
1)
(
)
(
)
2
2
;
-
+
+
2
2
4
a b
a
ab b
2)
(
)
(
)
2
2
9
6
4
;
-
+
+
2
3
a
b
a
ab
b
3)
(
)
(
)
;
-
2
2
5
3
25
15
+ 9
x + y
x
xy
y
4)
(
)
(
)
2
6
4
.
ab
b
-
+
2
3
2
9
a+ b
a
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