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

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

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

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

=(x-y)(y-5)

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

tích mình với

ai tích mình 

mình tích lại

thanks nhiều

a) = (x - 1)(5x - 3)

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

= (x + y)(x - 5)

c) = x(x - y) - y(x - y)

= (x - y)^2

d) = 2(x² + 2x + 1 - y²)

= 2[(x + 1)² - y²]

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

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

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

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

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

a)5x-20y

=5x-4*5y

=5(x-4y)

b)x(x-1)-3x(x-1)

=(x-3x)(x-1)

c)x(x+y)-5x-5y

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

=(x-5)(x+y)

5.(x-4)

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

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

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 )

2x^2-5x-7

=2x^2+2x-7x-7

=2x.(x+!) -7.(x+1)

(2x-7).(x+1)

x^2-y^2-5x+5y

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

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

5x^3-5x^2.y-10x^2+10xy

=5x^2.(x-y)-10x.(x-y)

=(5x^2-10x).(x-y)

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

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)