a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

=3x (y-1-x)(y-1+x)

b) 3x² +11x+6

= 3 x² +9x +2x +6

=3x (x+3)+2(x+3)

= (x+3)(3x+2)

c) -x³ - 4xy² + 4x²y +16x

= -x (x² + 4y² - 4xy -16 )

= -x [(x -2y)² - 4² ]

= -x(x-2y-4)(x-2y+4) 

1.Ta có: 3x(x - y) - (y - x)² = 3x(x - y) - (x - y)² = (3x - x + y)(x - y) = (2x + y)(x - y)

2. Ta có: 2x(1 - 2x) + (2x - 3)(2x + 3) = 5

=> 2x - 4x² + 4x² - 9 = 5

=> 2x = 5 + 9

=> 2x = 14

=> x = 14 : 2 = 7

\(x^3+2x^2-2x-12=x^3-2x^2+4x^2-8x+6x-12\)

\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(x^2+4x+6\right)\)

\(x^3+2x^2-2x-12\)

\(=x^3-2x^2+4x^2-8x+6x-12\)

\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+4x+6\right)\)

hk tốt

^^

Ta có:\(25x^2-y^2+6yz-9z^2=25x^2-\left(y-3z\right)^2=\left(5x+y-3z\right)\left(5x-y+3z\right)\)

\(25x^2-y+6yz-9z^2\)

\(=\left(5x\right)^2-\left(y^2-6yz+9z^2\right)\)

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

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

Vậy \(25x^2-y^2+6yz-9z^2=\left(5x-y+3z\right)\left(5x+y-3z\right)\)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x^2-1\right)\left(x-3\right)^2=\left(x-1\right)\left(x+1\right)\left(x-3\right)^2\)