Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\\\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\\\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\\\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\). Thay vào \(y+z+1=2x\) ta được \(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)
+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\). Thay vào \(x+z+2=2y\) ta được \(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}+z=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}-\dfrac{4}{3}=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=\dfrac{-5}{6}\)