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

x² - y - y² - x

= (x² - y²) - x - y

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

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

\(x^2-2xy+y^2-z^2\\=(x^2-2xy+y^2)-z^2\\=(x-y)^2-z^2\\=(x-y-z)(x-y+z)\)

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

\(=3^2-\left(x-y\right)^2=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]=\left(3-x+y\right)\left(3+x-y\right)\)

\(9-x^2+2xy-y^2\)

\(=9-\left(x^2-2xy+y^2\right)\)

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

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

\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

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

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)\)

\(=\left(x+y-4\right)\left(x+y+3\right)\)

Phân tích thành nhân tử ak

\(=\left(y+x-4\right)\left(y+x+3\right)\)

= (x + y)² - ( x + y) - 12

= ( x + y ) ( x + y - 1) - 12

y² hay y ạ??? 

x² - 5x + 5

= x² - 5(x + 1)

= x² - \(\left[\sqrt{5\left(x+1\right)}\right]^2\)

= [x - \(\sqrt{5\left(x+1\right)}\)][x + \(\sqrt{5\left(x+1\right)}\)]

a)x²-2xy+y²-2x+2y

=(x²-2xy+y²-2x+2y+1)-1

=(x-y-1)²-1

=(x-y-2)(x-y)

b)x³-49x

=x(x²-7²)

=x(x+7)(x-7)

c)x²-y²+6x+9

=(x²+6x+9)-y²

=(x+3)²-y²

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

d)x²-6x+5

=x²-5x-x+5

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

=(x-1)(x-5)