\(\frac{2x+2y-z}{z}=\frac{2x-y+2z}{y}=\frac{-x+2y+2z}{x} \)

=>\(\frac{2x+2y-z}{z}+3=\frac{2x-y+2z}{y}+3=\frac{-x+2y+2z}{x}+3\)

=>\(\frac{2x+2y+2z}{z}=\frac{2x+2y+2z}{y}=\frac{2x+2y+2z}{x}\)

=>\(\frac{x+y+z}{z}=\frac{x+y+z}{y}=\frac{x+y+z}{x}\)

=>\(\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)

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 \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)

​Xin lỗi vì mik chưa học

bạn ấy xin lỗi thì mình cũng nói là tớ chưa học hihi

cái này dành cho lớp 8 hay lớp 7

TH1: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(y+z=-x\)

\(x+z=-y\)

\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)

TH2: \(x+y+z\ne0\)

\(\Rightarrow2x+2y-z=3\)

\(\Rightarrow2x+2y=4z\)

\(\Rightarrow x+y=2z\)

\(x+z=2y\)

\(y+z=2x\)

\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)

Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)

Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)

Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)

\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)

Vậy \(M=1\)

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

☯