ГДЗ Алгебра 8 класс Мерзляк, Полонский, Якир

ГДЗ Алгебра 8 класс Мерзляк, Полонский, Якир

авторы: , , .
издательство: "Вентана-Граф"

Раздел:

ГДЗ учебник по алгебре 8 класс Мерзляк, Полонский, Якир. §22. Упражнения. Номер №757

Сократите дробь:
1) 4 a 2 9 2 a 2 9 a 18 ;
2) 2 b 2 7 b + 3 4 b 2 4 b + 1 ;
3) c 2 5 c 6 c 2 8 c + 12 ;
4) m 3 1 m 2 + 9 m 10 ;
5) x 2 16 32 4 x x 2 ;
6) 4 n 2 9 n + 2 2 + 9 n 5 n 2 .

reshalka.com

ГДЗ учебник по алгебре 8 класс Мерзляк, Полонский, Якир. §22. Упражнения. Номер №757

Решение 1

4 a 2 9 2 a 2 9 a 18
2 a 2 9 a 18 = 0
D = b 2 4 a c = ( 9 ) 2 4 2 ( 18 ) = 81 + 144 = 251 > 0
a 1 = b + D 2 a = 9 + 251 2 2 = 9 + 15 4 = 24 4 = 6
a 2 = b D 2 a = 9 251 2 2 = 9 15 4 = 6 4 = 1 , 5
2 a 2 9 a 18 = 2 ( a 6 ) ( a ( 1 , 5 ) ) = ( a 6 ) ( 2 a + 3 )
4 a 2 9 2 a 2 9 a 18 = ( 2 a 3 ) ( 2 a + 3 ) ( a 6 ) ( 2 a + 3 ) = 2 a 3 a 6
Ответ: 2 a 3 a 6

Решение 2

2 b 2 7 b + 3 4 b 2 4 b + 1
2 b 2 7 b + 3 = 0
D = b 2 4 a c = ( 7 ) 2 4 2 3 = 49 + 24 = 25 > 0
b 1 = b + D 2 a = 7 + 25 2 2 = 7 + 5 4 = 12 4 = 3
b 2 = b D 2 a = 7 25 2 2 = 7 5 4 = 2 4 = 0 , 5
2 b 2 7 b + 3 = 2 ( b 3 ) ( b 0 , 5 ) = ( b 3 ) ( 2 b 1 )
2 b 2 7 b + 3 4 b 2 4 b + 1 = ( b 3 ) ( 2 b 1 ) ( 2 b 1 ) 2 = b 3 2 b 1
Ответ: b 3 2 b 1

Решение 3

c 2 5 c 6 c 2 8 c + 12
c 2 5 c 6 = 0
D = b 2 4 a c = ( 5 ) 2 4 1 ( 6 ) = 25 + 24 = 49 > 0
c 1 = b + D 2 a = 5 + 49 2 1 = 5 + 7 2 = 12 2 = 6
c 2 = b D 2 a = 5 49 2 1 = 5 7 2 = 2 2 = 1
c 2 5 c 6 = ( c 6 ) ( c ( 1 ) ) = ( c 6 ) ( c + 1 )
c 2 8 c + 12 = 0
D = b 2 4 a c = ( 8 ) 2 4 1 12 = 64 + 48 = 16 > 0
c 1 = b + D 2 a = 8 + 16 2 1 = 8 + 4 2 = 12 2 = 6
c 2 = b D 2 a = 8 16 2 1 = 8 4 2 = 4 2 = 2
c 2 8 c + 12 = ( c 6 ) ( c 2 )
c 2 5 c 6 c 2 8 c + 12 = ( c 6 ) ( c + 1 ) ( c 6 ) ( c 2 ) = c + 1 c 2
Ответ: c + 1 c 2

Решение 4

m 3 1 m 2 + 9 m 10
m 2 + 9 m 10 = 0
D = b 2 4 a c = 9 2 4 1 ( 10 ) = 81 + 40 = 121 > 0
m 1 = b + D 2 a = 9 + 121 2 1 = 9 + 11 2 = 2 2 = 1
m 2 = b D 2 a = 9 121 2 1 = 9 11 2 = 20 2 = 10
m 2 + 9 m 10 = ( m 1 ) ( m ( 10 ) ) = ( m 1 ) ( m + 10 )
m 3 1 m 2 + 9 m 10 = ( m 1 ) ( m 2 + m + 1 ) ( m 1 ) ( m + 10 ) = m 2 + m + 1 m + 10
Ответ: m 2 + m + 1 m + 10

Решение 5

x 2 16 32 4 x x 2
32 4 x x 2 = x 2 4 x + 32 = 0
D = b 2 4 a c = ( 4 ) 2 4 ( 1 ) 32 = 16 + 128 = 144 > 0
x 1 = b + D 2 a = 4 + 144 2 ( 1 ) = 4 + 12 2 = 16 2 = 8
x 2 = b D 2 a = 4 144 2 ( 1 ) = 4 12 2 = 8 2 = 4
x 2 4 x + 32 = ( x ( 8 ) ) ( x 4 ) = ( x + 8 ) ( x 4 )
x 2 16 32 4 x x 2 = ( x 4 ) ( x + 4 ) ( x + 8 ) ( x 4 ) = x + 4 x + 8
Ответ: x + 4 x + 8

Решение 6

4 n 2 9 n + 2 2 + 9 n 5 n 2
4 n 2 9 n + 2 = 0
D = b 2 4 a c = ( 9 ) 2 4 4 2 = 81 32 = 49 > 0
n 1 = b + D 2 a = 9 + 49 2 4 = 9 + 7 8 = 16 8 = 2
n 2 = b D 2 a = 9 49 2 4 = 9 7 8 = 2 8 = 1 4
4 n 2 9 n + 2 = 4 ( n 2 ) ( n 1 4 ) = ( n 2 ) ( 4 n 1 )
2 + 9 n 5 n 2 = 5 n 2 + 9 n + 2 = 0
D = b 2 4 a c = 9 2 4 ( 5 ) 2 = 81 + 40 = 121 > 0
n 1 = b + D 2 a = 9 + 121 2 ( 5 ) = 9 + 11 10 = 2 10 = 0 , 2
n 2 = b D 2 a = 9 121 2 ( 5 ) = 9 11 10 = 20 10 = 2
5 n 2 + 9 n + 2 = 5 ( n ( 0 , 2 ) ) ( n 2 ) = 5 ( n + 0 , 2 ) ( n 2 ) = ( 5 n + 1 ) ( n 2 )
4 n 2 9 n + 2 2 + 9 n 5 n 2 = ( n 2 ) ( 4 n 1 ) ( 5 n + 1 ) ( n 2 ) = 4 n 1 5 n + 1 = 1 4 n 5 n + 1
Ответ: 1 4 n 5 n + 1