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\(\frac{2x+2y-z}{z}=\frac{2x-y+2z}{y}=\frac{-x+2y+2z}{x} \)
=>\(\frac{2x+2y-z}{z}+3=\frac{2x-y+2z}{y}+3=\frac{-x+2y+2z}{x}+3\)
=>\(\frac{2x+2y+2z}{z}=\frac{2x+2y+2z}{y}=\frac{2x+2y+2z}{x}\)
=>\(\frac{x+y+z}{z}=\frac{x+y+z}{y}=\frac{x+y+z}{x}\)
=>\(\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Với \(x+y+z=0\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)
\(\Rightarrow M=\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\frac{-xyz}{8xyz}=-\frac{1}{8}\)
Với \(x=y=z\)\(\Rightarrow M=\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\frac{2x.2y.2z}{8xyz}=\frac{8xyz}{8xyz}=1\)
TH1 : \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)
Th2 : \(x+y+z\ne0\)
\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)
\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)
\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)
\(\Leftrightarrow x=y=z\)
\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)
Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)
