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

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

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

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

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

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

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

y² hay y ạ??? 

a, = (xy-y^2) + (2x-2y) = y(x-y) + 2.(x-y) = (x-y).(y+2) 

b, = (x+y)^2 - 9 = (x+y-3).(x+y+3)

\(xy-y^2-2x-2y\)   

=(xy-y).(2x-2y)

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

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

\(a,x^2+6x+9\)

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

\(b,10x-25-x^2\)

\(=-\left(x^2-10x+25\right)\)

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

\(c,8x^3-\frac{1}{8}\)

\(=8x^3-\left(\frac{1}{2}\right)^3\)

\(=\left(8x-\frac{1}{2}\right)\left(64x^2+4x+\frac{1}{4}\right)\)

\(d,8x^3+12x^2+6xy^2+y^3\)

\(=2\left(4x^3+6x^2+3xy^2+\frac{1}{2}y^3\right)\)

hok tốt!

ai tra loi dung minh cho

=(x^2-2xy-y^2)-(3x-3y)

=(x-y)^2-3(x-y)

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

x^2-2xy+y^2-9z^2

=(x-y)^2-9z^2

=(x-y)^2-(3z)^2

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

\(16y^2-4x^2-12x-9=16y^2-\left(4x^2+12x+9\right)=\left(4y\right)^2-\left(2x+3\right)^2\)\(=\left[4y-\left(2x+3\right)\right]\left(4y+2x+3\right)=\left(4y-2x-3\right)\left(4y+2x+3\right)\)