5x² - 10xy + 5y² - 20z²

=5(x² - 2xy + y² - 4z²)

= 5[ (x² - 2xy + y²) - (2z)²]

= 5[(x-y)² - (2z)²]

= 5(x-y-2z)(x-y+2z)

5x²-10xy+5y²-20z²

=5(x²-2xy+y²-4z²)

=5[(x-y)²-(2z)²]

=5(x-y-2z)(x-y+2z)

5²y² + 10xy³ - 20xy²

= 5y.(5y + 2xy² - 4xy)

\(5x^3-5x^2y-10x^2+10xy=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)

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)

x⁴-x³-x²+1

=x³(x-1)-(x²-1)

=x³(x-1)-(x-1)(x+1)

=(x-1)(x³-x-1)

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)

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)