\(\text{ABTC dãy tỉ số bằng nhau , ta có:}\)
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\text{=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2y+2z+2x}{x+y+z}}\)\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\)\(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)\(=\frac{2y+2z+2x}{x+y+z}\)\(=\frac{2\left(x+y+z\right)}{x+y+z}\)\(=2\)
\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}y+z+1=2x\\x+z+2=2y\\y+z+1=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+z+y+2=3y\\z+x+y-3=3z\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+2=3y\\\frac{1}{2}-3=3z\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}\)
