TH1:x+y+z=0
\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(\Rightarrow M=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8xyz}=\frac{-xyz}{8xyz}=\frac{-1}{8}\)
TH2: \(x+y+z\ne0\)
Ta có:
\(\frac{2x+2y-z}{z}=\frac{2x-y+2z}{y}=\frac{-x+2y+2z}{x}\)
\(\Rightarrow\left(\frac{2x+2y-z}{z}+3\right)=\left(\frac{2x-y+2z}{y}+3\right)=\left(\frac{-x+2y+2z}{x}+3\right)\)\(\Rightarrow\frac{2x+2y+z}{z}=\frac{2x+2y++2z}{y}=\frac{2x+2y+2z}{x}\)
\(\Rightarrow x=y=z\)
\(\Rightarrow M=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8xyz}=\frac{2x.2y.2z}{8xyz}=1\)
Vậy M=1 hoặc M=\(\frac{-1}{8}\)
theo bài ra ta có:
\(\frac{2x+2y-z}{z}=\frac{2x-y+2z}{y}=\frac{-x+2y+2z}{x}\)
\(\Rightarrow\frac{2x+2y-z}{x}+3=\frac{2x-y+2z}{y}+3=\frac{2y+2z-x}{x}+3\)
\(\Rightarrow\frac{2x+2y+2z}{z}=\frac{2x+2z+2y}{y}=\frac{2y+2z+2x}{x}\)
vì x;y;z là các số hữu tỉ khác 0
=> x = y = z
vậy ta có:
\(M=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8xyz}=\frac{2x.2y.2z}{8xyz}=\frac{8xyz}{8xyz}=1\)
vậy M = 1
