Nếu a+b+c ≠≠ 0 thì theo tính chất dãy tỉ số bằng nhau :

a/(b+c)=b/(a+c) = c/(a+b) =(a+b+c)/2(a+b+c)=1/2

Nếu a+b+c = 0 thì b+c = -a, c+a = -b, a+b = -c nên mỗi tỉ số a/(b+c)=b/(a+c) =c/(a+b)= -1
 

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

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

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

\(\Rightarrow a=b=c\)

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

Áp dụng TCDTSBN ta có:

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

Nêu a+b+c khác 0 thi theo tinh chat day ti sô bang nhau ta co.                                     a/b+c=b/c+a=c/a+b=a+b+c/2(a+b+c)=1/2N êu a+b+c=0 thi b+c=-a; c+a=-b;a+b=-c. Nêna/b+c,b/c+a,c/a+b =-1

hay""'!!1!!!

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Vì =\(\frac{b}{a+c}\)=\(\frac{a}{b+c}\)\(\frac{c}{a+b}\)

Nên a=b=c

suy ra \(\frac{a}{b+c}\)=\(\frac{b}{a+c}\)=\(\frac{c}{a+b}\)

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