Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{y+x+1}=\frac{y}{x+z+1}=\frac{z}{y+z-2}=\frac{x+y+z}{2.\left(x+y+z\right)}=\frac{1}{2}\)
Hay x + y + z = \(\frac{1}{2}\)
\(\frac{x}{y+z+1}=\frac{1}{2}=>2x=y+z+1+=>3x=x+y+z+1=\frac{3}{2}\)
Tương tự tính y = 3/2
z = -3/2
\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+x+z+1+x+y-2}\) \(=\frac{x+y+z}{2\left(x+y+z\right)}\)
TH1: Nếu \(x+y+z=0\Rightarrow x=y=z=0\)
TH2: Nếu \(x+y+z\ne0\Rightarrow x+y+z=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
+) \(\frac{x}{y+z+1}=\frac{1}{2}\Rightarrow2x=y+z+1=\frac{1}{2}-x+1=\frac{3}{2}-x\)
\(\Rightarrow2x+x=\frac{3}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:3\Rightarrow x=\frac{1}{2}\)
+)\(\frac{y}{x+z+1}=\frac{1}{2}\Rightarrow2y=x+z+1=\frac{1}{2}-y+1=\frac{3}{2}-y\)
\(\Rightarrow2y+y=\frac{3}{2}\Rightarrow3y=\frac{3}{2}\Rightarrow y=\frac{3}{2}:3\Rightarrow y=\frac{1}{2}\)
+) \(\frac{z}{x+y-2}=\frac{1}{2}\Rightarrow2z=x+y-2=\frac{1}{2}-z-2=-\frac{3}{2}-z\)
\(\Rightarrow2z+z=\frac{-3}{2}\Rightarrow3z=\frac{-3}{2}\Rightarrow z=\frac{-3}{2}:3\Rightarrow z=\frac{-1}{2}\)
Vậy \(\left(x,y,z\right)=\left(0,0,0\right)\) hoặc \(\left(\frac{1}{2},\frac{1}{2},\frac{-1}{2}\right)\)
