x² - 2xy - 4z² + y²

= (x² - 2xy+y²) - (2z)²

= (x-y)²- (2z)²

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

x²-2xy-4z²+y²

=(x²-2xy+y²)-(2z)²

=(x-y)²-(2z)²

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

x^2 - 2xy - (4z)^2 + y^2

=(x-y)^2 - (4z)^2

= (x-y-4z)(x-y+4z)

x²-y²-2xy+y²

=x²-2xy

=x(x-2y)

x² - y² - 2xy + y²

= (x² - 2xy +y²) - y²

= (x-y)² - y²

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

= x(x-2y)

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

x² - y² + 2yz - z²

= x² - (y² - 2yz + z²)

= x² - (y-z)²

= (x-y+z)(x+y-z) 

Câu 1 : x²-y²+2yz-z²=-(y²-2yz+z²-x²)                    Câu 2: x²-2xy+y²-xz+yz=(x²-2xy+y²)-xz+yz

                                 =-(y-z)² -x²                                                                   =(x-y)²-z(x-y)

                                        =-(y-z-x)(y-z+x)                                                            =(x-y)(x-y-z)

x³ - 3x² - 4x + 12

= x²(x-3) - 4(x-3)

= (x-3)(x² -4)

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

\(x^3-3x^2-4x+12\)

\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x^2-4\right)\left(x-3\right)\)

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

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

x-y²+x²-y

=(x-y)+(x²-y²)

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

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

ta có :

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

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

x²-5x+5y-y²

=(x²-y²)-(5x-5y)

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

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

ta có :

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

x⁴-x³-x²+1

=x³(x-1)-(x²-1)

=x³(x-1)-(x-1)(x+1)

=(x-1)(x³-x-1)