\(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

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

\(5x^2+5xy+x+y\\ =5x\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(5x+1\right)\)

`5x^2 + 5xy +x +y`

`=(5x^2 + 5xy )+(x+y)`

`=5x(x+y)+(x+y)`

`=(x+y)(5x+1)`

 5\(x^2\)+5xy+x+y = (5\(x^2\)+x)+(5xy+y)

                       = x(5x+1)+y(5x+1)

                       = (5x+1)(x+y)

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

= (5xy-5x)-(y-1)

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

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

5x² - 5xy + 4y - 4x

= 5x ( x - y ) - 4 ( x - y )

= ( 5x - 4 ) ( x - y )

( x + y )³ + ( x - y )³

= 2x³ + 6xy²

= 2x ( x² + 3y² )

5 x^2 - 5xy + 4y - 4x 

= 5x ( x - y ) - 4 ( x - y ) 

= ( x - y ) ( 5x - 4 ) 

( x + y )^3 + ( x - y )^3 

= \(x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\) 

= \(2x^3+6xy^2\) 

=\(2x\left(x^2+3y^2\right)\) 

a) x( x + 2 )( x + 3 )( x + 5 ) + 5

= [ x( x + 5 ) ][ ( x + 2 )( x + 3 ) ] + 5

= ( x² + 5x )( x² + 5x + 6 ) + 5 (1)

Đặt t = x² + 5x

(1) <=> t( t + 6 ) + 5

       = t² + 6t + 5

       = t² + t + 5t + 5 

       = t( t + 1 ) + 5( t + 1 )

       = ( t + 1 )( t + 5 )

       = ( x² + 5x + 1 )( x² + 5x + 5 )

b) 6x² - 5xy + y² = 6x² - 3xy - 2xy + y² = 3x( 2x - y ) - y( 2x - y ) = ( 2x - y )( 3x - y )

a,\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+5\)

\(=x\left(x+5\right)\left(x+2\right)\left(x+3\right)+5\)

\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+5\)(*)

Đặt \(a=x^2+5x\)ta đc:

(*)=\(a\left(a+6\right)+5\)

\(=a^2+6a+5\)

\(=a^2+a+5a+5\)

\(=a\left(a+1\right)+5\left(a+1\right)\)

\(=\left(a+5\right)\left(a+1\right)\)

\(=\left(x^2+5x+5\right)\left(x^2+5x+1\right)\)

b,\(6x^2-3xy-2xy+y^2\)

\(=3x\left(2x-y\right)-y\left(2x-y\right)\)

\(=\left(3x-y\right)\left(2x-y\right)\)

x 2355

2x^2-5xy-3y^2

 = 2^x + xy - 6xy - 3y^2  

= x(2x + y) - 3y(2x + y)  

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