Áp dụng tc dãy tỉ số băng nhau ta có:
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+x}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow\left\{\begin{matrix}\frac{x}{y+z+1}=\frac{1}{2}\\\frac{y}{z+x+1}=\frac{1}{2}\\\frac{z}{x+y-2}=\frac{1}{2}\\\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}2x=y+z+1\\2y=z+x+1\\2z=x+y-2\end{matrix}\right.\) (1)
Mà \(x+y+z=\frac{1}{2}\Rightarrow\left\{\begin{matrix}y+z=\frac{1}{2}-x\\z+x=\frac{1}{2}-y\\x+y=\frac{1}{2}-z\end{matrix}\right.\) (*)
Thay (*) vào (1) ta được
\(\left\{\begin{matrix}2x=\frac{1}{2}-x+1\\2y=\frac{1}{2}-y+1\\2z=\frac{1}{2}-z-2\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}3x=\frac{3}{2}\\3y=\frac{3}{2}\\3z=\frac{-3}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{matrix}\right.\)
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