Question

tính. x^2\(y^2+z^2-x^2)+y^2\(z^2+x^2-y^2)+z^2\(x^2+y^2-z^2) cho x+y+z=0, x,y,zkhac 0

Answer 1

\(x^2\left(y^2+z^2-x^2\right)+y^2\left(z^2+x^2-y^2\right)+z^2\left(x^2+y^{ 2}-z^2\right)\)

\(=x^2\left[\left(y+z\right)^2-x^2-2yz\right]+y^2\left[\left(z+x\right)^2-y^2-2zx\right]+z^2\left[\left(x+y\right)^2-z^2-2xy\right]\)

\(=x^2\left[\left(y+z-x\right)\left(y+z+x\right)-2xy\right]+y^2\left[\left(z+x-y\right)\left(z+x+y\right)-2zx\right]\)

\(+z^2\left[\left(x+y-z\right)\left(x+y+z\right)-2xy\right]\)

\(=x^2\left[\left(y+z-x\right).0-2yz\right]+y^2\left[\left(z+x-y\right).0-2zx\right]+z^2\left[\left(x+y-z\right).0-2xy\right]\)

\(=x^2\left(-2yz\right)+y^2\left(-2zx\right)+z^2\left(-2xy\right)\)\(=-2x^2yz-2xy^2z-2xyz^2\)

\(=-2xyz\left(x+y+z\right)=-2xyz.0=0\)