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

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

Học tốt

Ta có : x² - y² - 5x - 5y

= ( x² - y² ) - ( 5x + 5y )

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

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

cảm ơn bạn moon nhé bạn đã giúp mình rất rất nhiều ( nick trên cx là cảu mình )

\(5x^2+3\left(x+y\right)-5y^2=5\left(x+y\right)\left(x-y\right)+3\left(x+y\right)=\left(x+y\right)\left[5\left(x-y\right)+3\right]\)

\(a,=x^2-9-x^2+6x-9=6x-18\\ b,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

Câu 1 đề sai rồi bạn ơi 

Câu 2/ 5(x² + 2xy + y² ) = 5(x +y )²

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

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

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

                                        \(=\left(x-y\right)\cdot\left[\left(x+y\right)-5\right]=\left(x-y\right)\cdot\left(x+y-5\right)\)

a)Ta có:  \(x\left(x-y\right)+5x-5y=x\left(x-y\right)+\left(5x-5y\right)\)

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

b) \(x^2-y^2+10x+25=\left(x^2+10x+25\right)-y^2=\left(x+5\right)^2-y^2\)

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

x² - 5x + 5y - y² =

= -5 (x-y) + (x² - y² )

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

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

x²-5x+5y-y²

=x²-y²-5x+5y

=(x²-y²)-5(x+y)

=(x²-y²)(-5+x+y)