Ta có :
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\dfrac{2a+b+c+d-a}{a}=\dfrac{a+2b+c+d-b}{b}=\dfrac{a+b+2c+d-c}{c}=\dfrac{a+b+c+2d-d}{d}\)
\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1}{d}\)
\(\Rightarrow a=b=c=d\)
\(M=\dfrac{c+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(=\dfrac{c+c}{c+c}+\dfrac{c+c}{c+c}+\dfrac{c+c}{c+c}+\dfrac{c+c}{c+c}\)
\(=1+1+1+1\\ =4\)
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