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 :\

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

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

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

Ta có :   2a + b + c+ d / a - 1 = a + 2b + c + d / b - 1 = a + b + 2c + d / c - 1 = a + b + c +2d / d - 1

  => a + b + c + d / a =  a + b + c + d / b = a + b + c + d / c = a + b + c + d / d

Xét 2 trường hợp : 

TH1:   a + b + c + d = 0

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

Khi đó M = ( -1 ) . 4 = -4

TH2 :  a + b + c + d  khác 0 

=> a = b = c = d

Khi đó M = 1 . 4 = 4

Vậy M = 4 hoặc M = - 4

bn lik e bai mjh dc ko 

2a+b+c+d/a=a+2b+c+d/b=a+b+2c+d/c=a+b+c+2d/d  

\(\Leftrightarrow\)a+a+b+c+d/a=a+b+b+c+d/b=a+b+c+c+d/c=a+b+c+d+d/d  

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

\(\Leftrightarrow\)a+b+c+d/a=a+b+c+d/b=a+b+c+d/c=a+b+c+d/d  

Dến đây ta xét 2 TH:  

a+b+c+d≠0  

a+b+c+d=0

Ta có: (2a+b+c+d/a)-1=(a+2b+c+d/b)-1=(a+b+2c+d/d)-1=(a+b+c+2d/d)-1

=>a+b+c+d/a=a+c+b+d/b=a+b+c+d/c=a+b+c+d/d

=>a=b=c=d         (1)

Ta lại có: a+b+c+d/a=a+b+c+d/b=a+b+c+d/c=a+b+c+d/d

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

=4(a+b+c+d)/a+b+c+d=4          (2)

Từ (1) và (2) => a=b=c=d=1

=> M=1+1/1+1 + 1+1+/1+1 + 1+1/1+1 + 1+1/1+1

=4

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

>a=b=c=d>tự tính