theo bài ra ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\) \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\) \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\) vì a; b;c ; d khác 0
=> a = b = c = d
=> \(M=1+1+1+1=4\)
vậy M = 4
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}=\dfrac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{{}\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
\(\Rightarrow M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy \(M=4\)