\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
Áp dụng t/c dãy tỉ số bằng nhau , ta có :
\(\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{x+y-3}{z}\)
= \(\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)
= \(\frac{2\left(x+y+z\right)}{x+y+z}=2\)
+) \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}=0,5\)
\(\Rightarrow\left\{\begin{matrix}x+y=0,5-z\\x+z=0,5-y\\y+z=0,5-x\end{matrix}\right.\)
=> \(\frac{0,5-x+1}{x}=\frac{0,5-y+2}{y}=\frac{0,5-z-3}{z}\)
\(\Rightarrow\left\{\begin{matrix}\frac{1,5-x}{x}=2\\\frac{2,5-y}{y}=2\\\frac{-2,5-z}{z}=2\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}1,5-x=2x\\2,5-y=2y\\-2,5-z=2z\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}1,5=3x\\2,5=3y\\-2,5=3z\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=0,5\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{matrix}\right.\)
