a) Ta có: \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
b) Ta có: \(12xy-12xz+3x^2y-3x^2z\)
\(=3x\left(4y-4z+xy-xz\right)\)
\(=3x\left[4\left(y-z\right)+x\left(y-z\right)\right]\)
\(=3x\left(y-z\right)\left(4+x\right)\)
c) Ta có: \(\frac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\frac{1}{2}\left(x^2+y^2\right)^2-\frac{1}{2}\cdot4x^2y^2\)
\(=\frac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\)
\(=\frac{1}{2}\left[x^4+2x^2y^2+y^4-4x^2y^2\right]\)
\(=\frac{1}{2}\left(x^4-2x^2y^2+y^4\right)\)
\(=\frac{1}{2}\left(x^2-y^2\right)^2\)
\(=\frac{1}{2}\left(x-y\right)^2\cdot\left(x+y\right)^2\)
d) Ta có: \(-50x^2y^2+2\left(x-y\right)^2\)
\(=-2\left[25x^2y^2-\left(x-y\right)^2\right]\)
\(=-2\left(25x^2y^2-x+y\right)\left(25x^2y^2+x-y\right)\)
a) \(x^3+x^2y-x^2z-xyz\)
\(=x\left(x^2+xy-xz-yz\right)\)
\(=x\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=x\left(x+y\right)\left(x-z\right)\)
b) \(12xy-12xz+3x^2y-3x^2z\)
\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)
\(=\left(y-z\right)\left(12x+3x^2\right)\)
c) \(\frac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\frac{1}{2}\left(x^2+y^2\right)^2-\frac{1}{2}.4x^2y^2\)
\(=\frac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\)
\(=\frac{1}{2}\left[\left(x^2+y^2\right)^2-\left(2xy\right)^2\right]\)
\(=\frac{1}{2}\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)\)
\(=\frac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\)
d) \(-50x^2y^2+2\left(x-y\right)^2\)
\(=-2\left[25x^2y^2-\left(x-y\right)^2\right]\)
\(=-2\left[\left(5xy\right)^2-\left(x-y\right)^2\right]\)
\(=-2\left(5xy-x+y\right)\left(5xy+x-y\right)\)
P/s: Ko chắc ạ!
