Đặt \(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z=k\)

Áp dụng TC DTSBN ta có :

\(k=\frac{x+y+z}{\left(y+z+1\right)+\left(z+x+1\right)+\left(x+y-2\right)}=\frac{\left(x+y+z\right)}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\z+x+1=2y\\x+y-2=2z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+1=3y\\\frac{1}{2}-2=3z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)

ta có:\(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{y+x-2}=\frac{x+y+z}{2\left(x+y+x\right)}=\frac{1}{2}\)

\(\frac{y+z+1+x+z+1+x+y-3}{x+y+z}\)=\(\frac{2\left(X+Y+Z\right)}{x+y+z}\)=2  =>x+y+z=\(\frac{1}{2}\)   tu lam di nhe