\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
=>\(1+\frac{a+b+c+d}{a}=1+\frac{a+b+c+d}{b}=1+\frac{a+b+c+d}{c}=1+\frac{a+b+c+d}{d}\)
=>\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
=>a=b=c=d
=>a+b=c+d; b+c=d+a
=>M=1+1+1+1=4
ta có :
\(\frac{2a+b+c+d}{a}-1=\frac{a+b+c+d}{a}\)
\(\frac{a+2b+c+d}{b}-1=\frac{a+b+c+d}{b}\)
\(\frac{a+b+2c+d}{c}-1=\frac{a+b+c+d}{c}\)
\(\frac{a+b+c+2d}{d}-1=\frac{a+b+c+d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}\)
+) Nếu \(a,b,c,d\ne0\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
+) Nếu \(a+b+c+d=0\Rightarrow a+b=-\left(c-d\right);b+c=-\left(d+a\right)\)
\(\Rightarrow M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
ở đây họ ko ra điều kiện j thì mình nghĩ xét 2 TH
