Ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}.\)
\(\Rightarrow\frac{a+b+c+d}{a}+\frac{a}{a}=\frac{a+b+c+d}{b}+\frac{b}{b}=\frac{a+b+c+d}{c}+\frac{c}{c}=\frac{a+b+c+d}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}+1=\frac{a+b+c+d}{b}+1=\frac{a+b+c+d}{c}+1=\frac{a+b+c+d}{d}+1\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}.\)
Từ đây ta xét 2 trường hợp sau:
+ TH1: \(a+b+c+d=0.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
Ta có: \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(\Rightarrow M=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(d+a\right)}{d+a}+\frac{-\left(a+b\right)}{a+b}+\frac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow M=-4.\)
+ TH2: \(a+b+c+d\ne0.\)
\(\Rightarrow a=b=c=d.\)
Lại có: \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4.\)
Vậy \(M=-4\) hoặc \(M=4.\)
Chúc bạn học tốt!
