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

x-y²+x²-y

=(x-y)+(x²-y²)

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

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

x⁴-x³-x²+1

=x³(x-1)-(x²-1)

=x³(x-1)-(x-1)(x+1)

=(x-1)(x³-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)

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)

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

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

=(x+y)(x-5)

x³ - 3x² - 4x + 12

= x²(x-3) - 4(x-3)

= (x-3)(x² -4)

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

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

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

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

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

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

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)