Xét 2 trường hợp :
TH1 : \(x=y=z=0\) (thõa mãn)
TH2 : \(x;y;z\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(x+y+z=\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\\dfrac{x}{y+z-2}=\dfrac{1}{2}\\\dfrac{y}{z+x+5}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-5}{6}\\z=\dfrac{11}{6}\end{matrix}\right.\)
Vậy ...
Theo đề bài ta có:
\(\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z-2}=\dfrac{y}{z+x-3}=\dfrac{z}{x+y+5}\)
\(=\dfrac{x+y+z}{y+z-2+z+x-3+x+y+5}\)
\(=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}\)
Ta có:
\(\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\\dfrac{x}{y+z-2}=\dfrac{1}{2}\\\dfrac{y}{z+x-3}=\dfrac{1}{2}\\\dfrac{z}{x+y+5}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\2x=y+z-2\\2y=z+x-3\\2z=x+y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\dfrac{1}{2}\\y+z-2x=2\\x+z-2y=3\\2z-x-y=5\end{matrix}\right.\)
Cạn r ......
