Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{y+z-2}{x+1}=\frac{z+x+1}{y-1}=\frac{x+y-3}{z-2}=\frac{y+z-2+z+x+1+x+y-3}{x+1+y-1+z-2}=\frac{2x+2y+2z-4}{x+y+z-2}=2\)
=>\(\begin{cases}y+z-2=2\left(x+1\right)\\ z+x+1=2\left(y-1\right)\\ x+y-3=2\left(z-2\right)\end{cases}\Rightarrow\begin{cases}y+z=2x+4\\ x+z=2y-3\\ x+y=2z-1\end{cases}\)
Ta có: \(\frac{y+z-2}{x+1}=\frac{z+x+1}{y-1}=\frac{x+y-3}{z-2}=\frac{1}{x+y+z-2}\)
=>\(\frac{1}{x+y+z-2}=2\)
=>\(x+y+z-2=\frac12\)
=>\(x+y+z=\frac52\)
Ta có: \(x+y+z=\frac52\)
=>\(x+2x+4=\frac52\)
=>\(3x=\frac52-4=-\frac32\)
=>\(x=-\frac12\)
Ta có: \(x+y+z=\frac52\)
=>\(y+2y-3=\frac52\)
=>\(3y=\frac52+3=\frac{11}{2}\)
=>\(y=\frac{11}{6}\)
Ta có: \(x+y+z=\frac52\)
=>\(z+2z-1=\frac52\)
=>\(3z=\frac52+1=\frac72\)
=>\(z=\frac76\)
