TH1: x+y+z=0
\(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\z+x=-y\end{matrix}\right.\Rightarrow\dfrac{x+y}{z}=\dfrac{y+z}{x}=\dfrac{x+z}{y}=-1\)
TH2:\(x+y+z\ne0\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{x+y}{z}=\dfrac{y+z}{x}=\dfrac{z+x}{y}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)Vậy.....