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

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

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

\(\Rightarrow a=b=c\)

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

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

TH1: \(a+b+c+d=0\)

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

TH2: \(a+b+c+d\ne0\)

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

+) Nếu \(a,b,c\ne0\) thì theo tính chất dãy tỉ số bằng nhau ta có :

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

+) Nếu \(a+b+c=0\)

\(\Leftrightarrow b+c=-a;c+a=-b;a+b=-c\)

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

sửa lại đề bài nhé 

tìm x ,biết 

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

+ nếu a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{c}{a+b}\\\dfrac{a}{b+c}\\\dfrac{b}{c+a}\end{matrix}\right.\Rightarrow x=-1\)

nếu a+b+c \(\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có 

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

nếu nếu a+b+c \(\ne0\)

thì x=\(\dfrac{1}{2}\)

nếu nếu a+b+c =0

thì x= -1

x là giá trị của mỗi tỉ số nhé

\(\ne0\)\(\ne0\)

a+b+c=0\(\Rightarrow c=0-a-b\)

Thay vào tỉ số đầu tiên ta được

\(\dfrac{a}{b+c}=\dfrac{a}{b+0-a-b}=\dfrac{a}{-a}=-1\)

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

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

Chúc bạn học tốt

Ta có:

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

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

    \(=\dfrac{a+b+c+2d}{d}-1\)

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

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

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

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

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

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

\(\text{Trường hợp 1:}\)

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

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

\(\text{Trường hợp 2:}\)

\(\text{Nếu }a+b+c+d=0\text{ thì:}\)

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

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

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

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

*Nếu \(a+b+c\ne0\) \(\Rightarrow\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)*Nếu \(\) \(a+b+c=0\)

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

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

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

Theo tính chất của dãy tỉ số bằng nhau :

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

Vậy giá trị mỗi tỉ số trên là : \(\dfrac{1}{2}\)