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\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
Suy ra
\(x+y+z=\frac{1}{2}\)(1)
\(y+z+1=2x\)(2)
\(x+z+2=2y\)(3)
\(x+y-3=2z\)(4)
(2)-(1) ta có
\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)
\(y=-z\)
\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)
\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)
\(z=-\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)