\(a,5x^2y-10xy^2=5xy\left(x-2y\right)\\ b,x^2+2xy+y^2-5x-5y=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\\ c,x^2-6x+8=\left(x^2-2x\right)-\left(4x-8\right)=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\\ d,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)

5x^2+10xy+5y^2

=5.(x²+2xy+y²)

=5.(x+y)²

x^3-6x^2+9x

=x.(x²-6x+9)

=x.(x-3)²

xy+y^2-x-y

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

=(x+y)(y-1)

haha bạn ấy nhấn  đúng cho tui nè

1)3x⁴-5x³y+6x²-10xy+2

=(3x⁴-5x³y)+(6x²-10xy)+2

=x³(3x-5y)+2x(3x-5y)+2

=x³.0+2x.0+2

=0+0+2

=2

2) x⁵-2010x⁴+2010x³-2010x²+2010x-2020

=x⁵-(2009+1)x⁴+(2009+1)x³-(2009+1)x²+(2009+1)x-2009-11

=x⁵-(x+1)x⁴+(x+1)x³-(x+1)x²+(x+1)x-x-11

=x⁵-x⁵-x⁴+x⁴+x³-x³-x²+x²+x-x-11

=-11

2, Với x= 2009 => 2010=x+1

=> \(x^5-2010\text{x}^4+2010\text{x}^3-2010\text{x}^2+2010\text{x}-2020=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2020\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2020\)

\(=x-2020\)

\(=2009-2020\\ =-11\)

a/ = xy(5y - 6x)

b/ = - (15y² - 2x + 5y - 10xy)

a. 5(x^2-2xy+y^2-4z^2)=5[(x-1)^2-(2z)^2]=5(x-1-2z)(x-1+2z)

b.6x^2-23x-18=6^2-4x+27x-18= 2x(3x-2)+9(3x-2)=(2x+9)(3x-2)

a) 2x-5y+4y+2x

=4x+y

Tai x=3 y=-12 thi

4x3+(-12)=12-12=0

b)3x+4y-2x-3y

b)3x+4y-2x-3y

=x+y

Tai x-2; y=-5 thi

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