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Theo đề, ta có: \(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=\dfrac{t}{x}\) \(=\dfrac{x+y+z+t}{y+z+t+x}=1\) .
\(\Rightarrow x=y;y=z;z=t;t=x\)
\(\Rightarrow x=y=z=t\)
\(M=\dfrac{2x-y}{z+t}+\dfrac{2y-z}{t+x}+\dfrac{2z-t}{x+y}+\dfrac{2t-x}{y-z}\)
\(M=\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}\)
\(M=\dfrac{1}{2}.4\)
\(M=2\)
Ta có
\(\frac{2x+y+z+t}{x}=\frac{x+2y+z+t}{y}=\frac{x+y+2z+t}{z}=\frac{x+y+z+2t}{t}\)
\(\Rightarrow1+\frac{x+y+z+t}{x}=1+\frac{x+y+z+t}{y}=1+\frac{x+y+z+t}{z}=1+\frac{x+y+z+t}{t}\)
\(\Rightarrow\frac{x+y+z+t}{x}=\frac{x+y+z+t}{y}=\frac{x+y+z+t}{z}=\frac{x+y+z+t}{t}\)
Xét 2 trường hợp
Nếu \(x+y+z+t=0\)
\(\Rightarrow\left\{\begin{matrix}x+y=-z-t\\y+z=-t-x\\t+x=-y-z\\z+t=-x-y\end{matrix}\right.\)
Ta có \(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(=\frac{-z-t}{z+t}+\frac{-t-x}{t+x}+\frac{-x-y}{x+y}+\frac{-y-z}{y+z}\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=\left(-4\right)\)
Nếu \(x=y=z=t\)
Ta có \(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}\)
\(=1+1+1+1\)
\(=4\)