Ta có: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z=\frac{x+y+z}{z+y+1+x+z+1+x+y-2}=\frac{x+y+z}{2x+2y+2z}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
✿ \(\frac{z}{x+y-2}=\frac{1}{2}\Rightarrow2z=x+y-2\Rightarrow2z+2=x+y\)
\(\text{Lại có: }x+y+z=\frac{1}{2}\Rightarrow2z+2+z=\frac{1}{2}\Rightarrow3z+2=\frac{1}{2}\Rightarrow3z=\frac{1}{2}-2=-\frac{3}{2}\Rightarrow z=\frac{-3}{2}:3=-\frac{1}{2}\)
✿ \(\frac{y}{x+z+1}=\frac{1}{2}\Rightarrow2y=x+z+1\Rightarrow2y-1=x+z\)
\(\text{Lại có: }x+y+z=\frac{1}{2}\Rightarrow2y-1+y=\frac{1}{2}\Rightarrow2y+y=\frac{1}{2}+1=\frac{3}{2}\Rightarrow3y=\frac{3}{2}\Rightarrow y=\frac{3}{2}:3=\frac{1}{2}\)
\(\text{✿}x+y+z=\frac{1}{2}\Rightarrow x+\frac{1}{2}+\left(-\frac{1}{2}\right)=\frac{1}{2}\Rightarrow x+0=\frac{1}{2}\Rightarrow x=\frac{1}{2}-0=\frac{1}{2}\)