trừ mỗi tỉ lệ cho 1 ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
\(\Rightarrow\frac{2a+b+c+d}{a}-\frac{a}{a}=\frac{a+2b+c+d}{b}-\frac{b}{b}=\frac{a+b+2c+d}{c}-\frac{c}{c}=\frac{a+b+c+2d}{d}-\frac{d}{d}\)
\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
+Nếu a+b+c+d\(\ne\)0 thì a=b=c=d lúc đó
M=1+1+1+1=4
+Nếu a+b+c+d=0 thì a+b=-(c+d);b+c=-(d+a);c+d=-(a+b);d+a=-(b+c) lúc đó:
M=(-1)+(-1)+(-1)+(-1)=-4
\(\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{a+b+2c+d+a+b+c+2d}{c+d}=\frac{2a+2b+3c+3d}{c+d}\)
\(=\frac{2\left(a+b\right)}{c+d}+\frac{3\left(c+d\right)}{c+d}=2.\frac{a+b}{c+d}+3\)
\(\frac{2a+b+c+d}{a}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+b+c+2d}{a+d}=\frac{3a+3d+2c+2b}{a+d}\)
\(=\frac{3\left(a+d\right)}{a+d}+\frac{2\left(b+c\right)}{a+d}=3+2.\frac{b+c}{a+d}\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{2a+b+c+d+a+2b+c+d}{a+b}=\frac{3a+3b+2c+2d}{a+b}\)
\(=\frac{3\left(a+b\right)}{a+b}+\frac{2\left(c+d\right)}{a+b}=3+\frac{c+d}{a+b}.2\)
\(\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+2b+c+d+a+b+2c+d}{b+c}=\frac{3b+3c+2a+2d}{b+c}\)
\(=\frac{3\left(b+c\right)}{b+c}+\frac{2\left(a+d\right)}{b+c}=3+\frac{a+d}{b+c}.2\)
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)
\(\Rightarrow\frac{2a+b+c+d}{a}+\frac{a+2b+c+d}{b}+\frac{a+b+2c+d}{c}+\frac{a+b+c+2d}{d}=5.4=20\)
\(\Rightarrow3+\frac{a+b}{c+d}.2+3+\frac{b+c}{a+d}.2+3+\frac{c+d}{a+b}.2+3+\frac{d+a}{b+c}.2=20\)
\(\Rightarrow2.\left(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\right)=20-3-3-3-3\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=8:2=4\)
vậy \(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4\)
