Lời giải:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
\(\Leftrightarrow 1+\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}+1=\frac{a+b+c+d}{c}+1=\frac{a+b+c+d}{d}+1\)
\(\Leftrightarrow \frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}(*)\)
Từ đây ta xét 2 TH sau:
TH1: \(a+b+c+d=0\)
\(\Rightarrow \left\{\begin{matrix} a+b=-(c+d)\\ b+c=-(d+a)\\ c+d=-(a+b)\\ d+a=-(b+c)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} \frac{a+b}{c+d}=-1\\ \frac{b+c}{d+a}=-1\\ \frac{c+d}{a+b}=-1\\ \frac{d+a}{b+c}=-1\end{matrix}\right.\)
\(\Rightarrow M=-4\)
TH2: \(a+b+c+d\neq 0\)
Khi đó từ $(*)$ suy ra a=b=c=d
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=4\)
Vậy $M=4$ hoặc $M=-4$
