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

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

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

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

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

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

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

\(=4\)

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

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

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

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

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

\(=-4\)

Ta có:

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

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

    \(=\dfrac{a+b+c+2d}{d}-1\)

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

Nếu a+b+c+d=0

⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)

Thay vào M, ta có:

\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)

Nếu a+b+c+d ≠0

⇒ \(a=b=c=d\)

Thay vào M, ta có

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

\(\text{Cùng trừ mỗi tỉ số trên 1 đơn vị ta được:}\)

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

\(\text{Từ đây ta suy ra 2 trường hợp:}\)

\(\text{Trường hợp 1:}\)

\(\text{Nếu }a+b+c+d\notin0\Rightarrow a=b=c=d\)

\(\Rightarrow M=1+1+1+1=1.4=4\)

\(\text{Trường hợp 2:}\)

\(\text{Nếu }a+b+c+d=0\text{ thì:}\)

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

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

\(\text{Do đó }M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Giải:

Ta có: \(\dfrac{2012a+b+c+d}{a}=\dfrac{a+2012b+c+d}{b}=\dfrac{a+b+2012c+d}{c}\)

\(=\dfrac{a+b+c+2012d}{d}\)

\(\Rightarrow\dfrac{2012a+b+c+d}{a}-2011=\dfrac{a+2012b+c+d}{b}-2011\)

\(=\dfrac{a+b+2012c+d}{c}-2011=\dfrac{a+b+c+2012d}{d}-2011\)

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

+) Xét \(a+b+c+d=0\) ta có:

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

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

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

+) Xét \(a+b+c+d\ne0\)

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

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

\(\Rightarrow M=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=\dfrac{2a}{2a}=1\)

Vậy nếu \(a+b+c+d=0\) thì M = -1

nếu \(a+b+c+d\ne0\) thì M = 1

Áp dụng dãy tỉ số bằng nhau :

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}=\dfrac{a+b+c+d}{b+c+d+a+c+d+a+b+d+b+c+a}=\dfrac{1}{3}\) \(\Rightarrow3a=b+c+d\left(1\right)\)

\(\Rightarrow3b=c+d+a\left(2\right)\)

\(\Rightarrow3c=a+b+d\left(3\right)\)

\(\Rightarrow3d=b+c+a\left(4\right)\)

Từ \(\left(1\right)+\left(2\right)\Rightarrow3a+3b=b+c+d+c+d+a\)

\(\Rightarrow2a+2b=2c+2d\)

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

Từ \(\left(2\right)+\left(3\right)\Rightarrow3b+3c=a+c+d+a+b+c\)

\(\Rightarrow2b+2c=2d+2a\)

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

Từ \(\left(1\right)+\left(3\right)\Rightarrow2a+2c=2b+2d\)

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

Ta có :

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

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

\(\Rightarrow a+b+2c=2a+a+d\)

\(\Rightarrow c=d\)

Lại có :

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

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

\(\Rightarrow2b+c+d=2a+d+c\)

\(\Rightarrow a=b\)

Từ những điều trên ta thấy được :

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

Nguyễn Thanh Hằng Xét thiếu TH rồi bạn !!!

Ta có :

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

TH1: Nếu a+b+c+d#0

thì Đỗ Thu Trà giải giống bạn Nguyễn Thanh Hằng

Nếu a+b+c+d=0 =>a+b=-(c+d); b+c=-(a+d);c+d=-(a+b); a+d=-(b+c)

Thế những cái này vao biểu thức M thì M=-4

Ta có:

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

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

+) Nếu \(a+b+c+d\ne0\) thì từ trên suy ra:\(a=b=c=d\)

\(\Rightarrow M=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)

+) Nếu \(a+b+c+d=0\)

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

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

Vậy M = 4 hoặc M = -4