Question

Rút gọn \(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\) biết x+y+z=0

Answer 1

Ta có: \(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(x^2+y^2+z^2+2xy+2yz+2xz=0\)

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

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

\(=\frac{-2.\left(xy+yz+zx\right)}{y^2+z^2+z^2+x^2+x^2+y^2-2.\left(xy+yz+zx\right)}\)

\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)}\)

\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left[-2.\left(xy+yz+zx\right)\right]-2.\left(xy+yz+zx\right)}\)

\(=\frac{-2.\left(xy+yz+zx\right)}{-6.\left(xy+yz+zx\right)}\)

\(=\frac{1}{3}\left(xy+yz+zx\ne0\right)\)

Tham khảo nhé~