\(=\left(a+b\right)\left(x+1\right)\)

\(b,=a\left(x+y\right)+2\left(x+y\right)=\left(a+2\right)\left(x+y\right)\\ c,=x\left(x+y\right)-2\left(x+y\right)=\left(x-2\right)\left(x+y\right)\\ d,=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)

từng câu 1 thôi:v

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

h: \(=\left(a-1-b\right)\left(a-1+b\right)\)

`ab(x-3) -a^2(x-3)`

`=(x-3)(a^2-ab)`

__

`ax+ay+bx+by`

`=a(x+y)+b(x+y)`

`=(x+y)(a+b)`

__

`ax+ay -2x-2y`

`=(ax+ay)-(2x+2y)`

`=a(x+y)-2(x+y)`

`=(x+y)(a-2_`

__

`2x-2y +ax-ay`

`=2(x-y)+a(x-y)`

`=(x-y)(2+a)`

__

`10ax -5ay -2x+y`

`= 5a(2x-y) -(2x-y)`

`=(2x-y)(5a-1)`

1: =(x-3)(ab-a^2)

=a(b-a)(x-3)

2: =a(x+y)+b(x+y)

=(x+y)(a+b)

3: =a(x+y)-2(x+y)

=(x+y)(a-2)

4: =2(x-y)+a(x-y)

=(x-y)(a+2)

5: =5a(2x-y)-(2x-y)

=(2x-y)(5a-1)

1) \(ab\left(x-3\right)-a^2\left(x-3\right)=a\left(x-3\right)\left(b-a\right)\)

2) \(ax+ay+bx+by=\left(a+b\right)\left(x+y\right)\)

3) \(ax+ay-2x-2y=\left(a-2\right)\left(x+y\right)\)

4) \(2x-2y+ax-ay=\left(2+a\right)\left(x-y\right)\)

5) \(10ax-5ay-2x+y=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(x-y\right)\)

\(2x^2-6xy+5x-15y=2x\left(x-3y\right)+5\left(x-3y\right)=\left(2x+5\right)\left(x-3y\right)\\ x^2-2x+1-4a^2=\left(x-1\right)^2-4a^2=\left(x-2a-1\right)\left(x+2a-1\right)\\ Sửa:25a^2-4x^2+4x-1=25a^2-\left(2x-1\right)^2=\left(5a-2x+1\right)\left(5a+2x-1\right)\\ 36x^2-a^2+10a-25=36x^2-\left(a-5\right)^2=\left(6x-a+5\right)\left(6x+a-5\right)\)

\(5a\left(2x-y\right)+\left(2x-y\right)=\left(5a+1\right)\left(2x-y\right)\)

Bài 4

c) x(x - 2) + (x - 2)²

= (x - 2)(x + x - 2)

= (x - 2)(2x - 2)

= 2(x - 2)(x - 1)

d) 2x(x - y)² - 5(y - x)

= 2x(x - y)² + 5(x - y)

= (x - y)(2x + 5)

Bài 5

a) x² - 6x - 2xy + 12y

= (x² - 6x) - (2xy - 12y)

= x(x - 6) - y(x - 6)

= (x - 6)(x - y)

b) 10ax - 5ay - 2x + y

= (10ax - 5ay) - (2x - y)

= 5a(2x - y) - (2x - y)

= (2x - y)(5a - 1)

c) x⁴ + x³y - x - y

= (x⁴ + x³y) - (x + y)

= x³(x + y) - (x + y)

= (x + y)(x³ - 1)

= (x + y)(x - 1)(x² + x + 1)

d) x³ + 2x² - 4x - 8

= (x³ + 2x²) - (4x + 8)

= x²(x + 2) - 4(x + 2)

= (x + 2)(x² - 4)

= (x + 2)(x + 2)(x - 2)

= (x + 2)²(x - 2)

e) xy - 5x - y² + 5y

= (xy - 5x) - (y² - 5y)

= x(y - 5) - y(y - 5)

= (y - 5)(x - y)

f) ax - bx - 2cx - 2a + 2b + 4c

= (ax - bx - 2cx) - (2a - 2b - 4c)

= x(a - b - 2c) - 2(a - b - 2c)

= (a - b - 2c)(x - 2)

g) 5x²y + 5xy² - b²x - b²y

= (5x²y + 5xy²) - (b²x + b²y)

= 5xy(x + y) - b²(x + y)

= (x + y)(5xy - b²)

h) 4x³ - 4x² - 9x + 9

= (4x³ - 4x²) - (9x - 9)

= 4x²(x - 1) - 9(x - 1)

= (x - 1)(4x² - 9)

= (x - 1)(2x - 3)(2x + 3)

6, \(x^2y+xy^2-4x-4y=xy\left(x+y\right)-4\left(x+y\right)=\left(xy-4\right)\left(x+y\right)\)

7, \(10ax-5ay-2x+y=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)

8, xem lại đề bạn nhé

9, \(4x^2-y^2+8y-16=4x^2-\left(y^2-8y+16\right)=4x^2-\left(y-4\right)^2\)

\(=\left(2x-y+4\right)\left(2x+y-4\right)\)

Trả lời:

6, x²y + xy² - 4x - 4y = ( x²y + xy² ) - ( 4x + 4y ) = xy ( x + y ) - 4 ( x + y ) = ( x + y )( xy - 4 )

7, 10ax - 5ay - 2x + y = ( 10ax - 5ay ) - ( 2x - y ) = 5a ( 2x - y ) - ( 2x - y ) = ( 2x - y )( 5a - 1 ) 

8, Sửa đề: x³ - 2x² + 2x - 4 = ( x³ - 2x² ) + ( 2x - 4 ) = x² ( x - 2 ) + 2 ( x - 2 ) = ( x - 2 )( x² + 2 )

9, 4x² - y² + 8y - 16 = 4x² - ( y² - 8y + 16 ) = 4x² - ( y - 4 )² = ( 2x - y + 4 )( 2x + y - 4 )

a ) -36a² + x² + 4y² - 4xy

= ( x² - 4xy + 4y² ) - (6a)²

= ( x -2y )² - (6a)²

= ( x - 2y - 6a ).(x - 2y + 6a )

b ) 10ax - 5ay +2x - y 

= ( 10ax - 5ay ) + ( 2x - y )

= 5a ( 2x - y ) + ( 2x - y )

= ( 2x - y ) . (5a + 1 ) 

c ) 2a²b(x + y) - 4ab²(-x - y )

= 2a²b( x+ y ) + 4ab²(x + y )

= 2ab(x + y ) ( a + 2b )

a, \(-36a^2+x^2+4y^2-4xy=\left(x+2y\right)^2-\left(6a\right)^2=\left(x+2y-6a\right)\left(x+2y+6a\right)\)

b, \(10ax-5ay+2x-y=5a\left(2x-y\right)+2x-y=\left(5a+1\right)\left(2x-y\right)\)

c, \(2a^2b\left(x+y\right)-4ab^2\left(-x-y\right)=2a^2b\left(x+y\right)+4ab^2\left(x+y\right)\)

\(=\left(2a^2b+4ab^2\right)\left(x+y\right)=2ab\left(a+2b\right)\left(x+y\right)\)

đầy đủ giúp em nhé