Áp dụng TCDTSBN ta có:
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{y+z+1}{x}=2\\\frac{x+z+2}{y}=2\\\frac{x+y-3}{z}=2\end{cases}\Rightarrow\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{cases}}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+2\\3z=\frac{1}{2}-3\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}}\)
Vậy x=1/2,y=5/6,z=-5/6