\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)(*)
\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}\)(Dãy tỉ số bằng nhau)
\(=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\Leftrightarrow x+y+z=\frac{1}{2}\)
Thay vào (*), ta có:
\(\frac{\left(\frac{1}{2}-x\right)+1}{x}=\frac{\left(\frac{1}{2}-y\right)+2}{y}=\frac{\left(\frac{1}{2}-z\right)-3}{z}=2\)
\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{2}-x\\2y=\frac{5}{2}-y\\2z=-\frac{5}{2}-z\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)
Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}.\)
