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

a/b=b/c=c/d=d/a=(a+b+c+d)/(b+c+d+a)=1

>a=b=c=d>tự tính

Ta có:\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d+a+2b+c+d+a+b+2c+d+a+b+c+2c}{a+b+c+d}=4\)

=>2a+b+c+d=4a

=>2a=b+c+d

Tương tự ta có:2b=a+c+d

2c=a+b+d

2d=a+b+c

=>2a+2b=b+c+d+a+c+d=>a+b+2c+2d

=>a+b=2c+2d

=>a+b/c+d=2

Tương tự ta có:b+c/d+a=2

c+d/a+b=2

d+a/b+c=2

=>M=2+2+2+2=8

Moon Light sai rồi bn nhé

Cộng vào bằng 5 nhé

\(TH1:a+b+c+d\ne0\)

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

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

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

\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)

\(=1+1+1+1\)

\(=4\)

\(TH2:a+b+c+d=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)

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

\(=-1-1-1-1\)

\(=-4\)

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

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

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

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

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

* TH1:  a + b + c + d \(\ne\)0

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

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

\(=\frac{5\left(a+b+c+d\right)}{a+b+c+d}=5\)

\(\Rightarrow2a+b+c+d=5a;a+2b+c+d=5b\)

\(\Rightarrow b+c+d=3a;a+c+d=3b\)

\(\Rightarrow b+c+d+a+c+d=3a+3b\)

\(\Rightarrow\left(a+b\right)+2\left(c+d\right)=3\left(a+b\right)\)

\(\Rightarrow2\left(c+d\right)=2\left(a+b\right)\)

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

CMTT ta được: \(b+c=a+d\)

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

* TH2: \(a+b+c+d=0\)

\(\Rightarrow a+b=-\left(c+d\right);b+c=-\left(d+a\right)\)

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

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