\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+2}=\dfrac{z}{x+y-3}=x+y+z=\dfrac{x+y+z}{2\left(x+y+z\right)}\)
TH1: \(x+y+z=0\Rightarrow x=y=z=0\)
TH2: \(x+y+z=\dfrac{1}{2}\) \(\Rightarrow\left\{{}\begin{matrix}y+z=\dfrac{1}{2}-x\\x+z=\dfrac{1}{2}-y\\x+y=\dfrac{1}{2}-z\end{matrix}\right.\)
\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\Rightarrow2x=y+z+1=\dfrac{1}{2}-x+1\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
\(\dfrac{y}{x+z+2}=\dfrac{1}{2}\Rightarrow2y=x+z+2=\dfrac{1}{2}-y+2\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
\(z=\dfrac{1}{2}-\left(x+y\right)=\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{5}{6}\right)=\dfrac{-5}{6}\)
Vậy \(\left(x;y;z\right)=\left(0;0;0\right)\) hoặc \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{5}{6};\dfrac{-5}{6}\right)\)
