\(=6x\left(-y^2+x^2+2x+1\right)\\ =6x\left[\left(x^2+2x+1\right)-y^2\right]\\ =6x\left[\left(x+1\right)^2-y^2\right]\\ =6x\left(x+1-y\right)\left(x+1+y\right)\)

\(\left(2x-y\right)^3=\left(2x\right)^3-3\left(2x\right)^2y+3\cdot2x\cdot y^2-y^3\\ =8x^3-12x^2y+6xy^2-y^3\)

Chọn \(12x^2y\)

\(10x\left(x-y\right)-6y\left(y-x\right)\)

\(=10x\left(x-y\right)+6x\left(x-y\right)\)

\(=\left(10x+6x\right)\left(x-y\right)\)

\(c,3x^2+5y-3xy-5x\)

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

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(3x-5\right)\left(x-y\right)\)

\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)

\(f,8x^3-12x^2y+6xy^2-y^3\)

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

\(g,x^3+8y^3=x^3+\left(2y\right)^3\)

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

\(i,x^2-25-2xy+y^2\)

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

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

Lời giải:
$A=(9x^2-6xy+y^2)+5y^2-6x-6y+20$

$=(3x-y)^2-2(3x-y)+4y^2-8y+20$

$=(3x-y)^2-2(3x-y)+1+(4y^2-8y+4)+15$

$=(3x-y-1)^2+(2y-2)^2+15\geq 15$

Vậy $A_{\min}=15$.

Giá trị này đạt tại $3x-y-1=2y-2=0$

$\Leftrightarrow (x,y)=(\frac{2}{3},1)$

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

e: \(=\left(x-2\right)\left(x-3\right)\)

a) x(4y-10x)

b)3(x+2y)+(x+1)

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

d)5x(y-z)²

e)(x-3)(x-2)

f)(2x+y)³