\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Ta có:
\(\dfrac{1}{x^2+2yz}=\dfrac{1}{x^2+yz-xy-xz}=\dfrac{1}{x\left(x-y\right)-z\left(x-y\right)}=\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(\dfrac{1}{y^2+2zx}=\dfrac{1}{y^2+zx-xy-yz}=-\dfrac{1}{\left(x-y\right)\left(y-z\right)}\)
\(\dfrac{1}{z^2+2xy}=\dfrac{1}{z^2+xy-yz-zx}=\dfrac{1}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}=\dfrac{\left(y-z\right)-\left(x-z\right)+\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
\(\Rightarrow...\)
