Áp dụng t/c dãy tỉ số bằng nhau ta có
[tex]\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2[/tex]
[tex]\Rightarrow \left\{\begin{matrix} 2x+2y+2z=1\\ 2x=y+z+1 \end{matrix}\right. \Rightarrow 2x=y+z+2x+2y+2z\Rightarrow y+z=0\Rightarrow x=\frac{1}{2}[/tex]
Từ [tex]\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=2\Rightarrow 2y=x+z+2\Rightarrow 3y=x+y+z+2=\frac{1}{2}+0+2=\frac{5}{2}\Rightarrow y=\frac{5}{6}\Rightarrow z=\frac{-5}{6}[/tex]