ĐK: ...
pt $(1)\Leftrightarrow \dfrac{x}{x-y}=-\dfrac{y}{y-z}-\dfrac{z}{z-x}=\dfrac{xy-2yz+z^2}{(y-z)(z-x)}$
$\Rightarrow \dfrac{x}{(x-y)^2}=\dfrac{xy-2yz+z^2}{(x-y)(y-z)(z-x)}$
cmtt: $\dfrac{y}{(y-z)^2}=\dfrac{yz-2zx+x^2}{(x-y)(y-z)(z-x)};\dfrac{z}{(z-x)^2}=\dfrac{zx-2xy+y^2}{(x-y)(y-z)(z-x)}$
$\Rightarrow \dfrac x{(x-y)^2}+\dfrac y{(y-z)^2}+\dfrac z{(z-x)^2}=\dfrac{x^2+y^2+z^2-xy-yz-zx}{(x-y)(y-z)(z-x)} \ (*)$
Từ $(*)$ và pt $(2)\Rightarrow \dfrac{x^2+y^2+z^2-xy-yz-zx}{(x-y)(y-z)(z-x)}=0$
$\Rightarrow x^2+y^2+z^2-xy-yz-zx=0$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0$
$\Leftrightarrow x=y=z$
=> vô nghiệm