Ta có\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

=> \(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

=> \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Khi a + b + c + d = 0

=> a + b = -(c + d)

b + c = -(a + d)

Khi đó \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)

\(=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{c+d}{-\left(c+d\right)}+\frac{a+d}{-\left(a+d\right)}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)= -4

Nếu a + b + d + d \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)

Khi đó M = \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{2a}{2a}+\frac{2b}{2b}+\frac{2c}{2c}+\frac{2d}{2d}=1+1+1+1=4\)

Vậy khi a + b + c + d = 0 => M = -4

khi a + b + c + d \(\ne\)0 => M = 4

Cho biểu thức sau:$\frac{2a+b+c+d}{a}$2 a + b + c + d a bam vao do nho bam lik e :\

\(\frac{a+b+c-2d}{a}=\frac{b+d+a-2c}{b}=\frac{b+d+c-2a}{c}=\frac{a+c+d-2b}{d}\)

\(=\frac{\left(a+b+c-2d\right)+\left(b+d+a-2c\right)+\left(b+d+c-2a\right)+\left(a+c+d-2b\right)}{a+b+c+d}\)

\(=\frac{a+b+c+d}{a+b+c+d}=1\)

\(\Leftrightarrow a=b=c=d\).

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{d}\right)\left(1+\frac{d}{a}\right)=2^4=16\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\left(\text{ vì a+b+c+d khác 0}\right)\)

\(\Rightarrow a=b=c=d\)

\(M=\frac{2a-b}{c+b}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2b-b}{b+b}+\frac{2c-c}{c+c}+\frac{2d-d}{d+d}=\frac{1}{2}.4=2\)