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

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

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

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

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

\(\text{Trường hợp 1:}\)

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

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

\(\text{Trường hợp 2:}\)

\(\text{Nếu }a+b+c+d=0\text{ thì:}\)

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

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

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

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

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

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

\(\Rightarrow a=b=c\)

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

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

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

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

*Nếu a+b+c+d=0

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

=> M=(-1)+(-1)+(-1)+(-1)=(-4)

Nếu a+b+c+d\(\ne\)0

=> a=b=c=d

=> M=1+1+1+1=4

Xét a+b+c+d=0

\(\Rightarrow\)a=-(b+c+d).Thay vào \(\dfrac{a}{b+c+d}\)ta có

\(\dfrac{-\left(b+c+d\right)}{b+c+d}\)=-1.Làm tương tự như thế ta có

M=-1+(-1)+(-1)+(-1)=-4

Xét a+b+c+d\(\ne\)0

Áp dụng tính chất dãy tỉ số bằng nhau ta có

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

=\(\dfrac{a+b+c+d}{2\cdot\left(a+b+c+d\right)}\)=\(\dfrac{1}{2}\)

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

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

\(\Rightarrow\)3a=a+b+c+d\(\left(1\right)\)

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

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

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

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

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

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

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

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

\(\Rightarrow\)3d=a+b+c+d\(\left(4\right)\)

Từ\(\left(1\right)\),\(\left(2\right)\),\(\left(3\right)\),\(\left(4\right)\)

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

\(\Rightarrow\)a=b=c=d.Khi đó

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

Vậy...

Mình trình bày hơi xấu các bạn thông cảm1!

ae tích nhiều vào nhe

Giải:

Ta có:

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c+d}=\dfrac{1}{3}\\\dfrac{b}{a+c+d}=\dfrac{1}{3}\\\dfrac{c}{a+b+d}=\dfrac{1}{3}\\\dfrac{d}{b+c+a}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a=b+c+d\\3b=a+c+d\\3c=a+b+d\\3d=b+c+a\end{matrix}\right.\Leftrightarrow a=b=c=d\)

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

\(\Leftrightarrow M=1+1+1+1=4\)

Vậy \(M=4\).

Chúc bạn học tốt!

Ta có

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

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

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

Trường hợp thứ nhất \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

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

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

\(\Rightarrow M=4\)

Trường hợp thứ hai\(a+b+c+d=0\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

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

\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(\Rightarrow M=-4\)

Vậy \(M\in\left\{4;-4\right\}\)

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

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

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

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

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

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

\(\Rightarrow M=4\)

Vậy .......

Chúc bạn học tốt!

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Bạn tự ghi điều kiện nha

Xét a+b+c+d=0

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

Khi đó M=-1+(-1)+(-1)+(-1)

=-4

có dãy tỉ số bằng nhau đó thì ta cộng vào rồi rút gọn thì được kết quả là \(\dfrac{2015}{2011}\) nó sẽ bằng với từng biểu thức đó.

Mẫu sẽ cố 2011=2011a; 2011=2011b; 2011=2011c; 2011=2011d

=> a = b = c = d = 1

=> M = 4