Question

[Trần Long]

 

Answer 1

[Nguyễn Việt Lâm]

- Với \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}=-1\\\dfrac{b}{c+a}=-1\\\dfrac{c}{a+b}=-1\end{matrix}\right.\)

- Với \(a+b+c\ne0\) áp dụng t/c dãy tỉ số bằng nhau:

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