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)

x² - 4x -5

= x² - 5x + x -5

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

=(x+1)(x-5)

x^2-4x-4=x²-2.2.x-2²

           =[x-2]²

\(x^2-4x-5=x^2-5x+x-5=\left(x^2-5x\right)+\left(x-5\right)=x\left(x-5\right)+\left(x-5\right)=\left(x+1\right)\left(x-5\right)\)

x⁴-x³-x²+1

=x³(x-1)-(x²-1)

=x³(x-1)-(x-1)(x+1)

=(x-1)(x³-x-1)

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)

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)

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

tiểu học mà bảo toán lớp 8

bạn đã suy nghĩ chưa

x^9-x^7-x^6-x^5+x^4+x^3+x^2-1
=>x^8(x+1)-x^7(x+1)-x^5(x+1)+x^3(x+1)+(x+1)(x-1)
<=>(x^8-x^7-x^5+x^3+x-1)(x+1)