Ta có: x+y+z=0
nên \(\left(x+y+z\right)^2=0\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz=0\)
Ta có: \(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{y^2-2xz+z^2+z^2-2xz+x^2+x^2-2xy+y^2}\)
\(=\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2xz-2yz}\)
\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2xz-2yz}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2xz+2yz\right)}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
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