Question

Cho \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}.\) Tính gtr của biểu thức : \(P=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)

Answer 1

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{-c}{c}+\dfrac{-a}{a}+\dfrac{-b}{b}=-3\)

TH2: \(a+b+c\ne0\)

Áp dụng t/c dãy tỉ số bằng nhau:

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}=\dfrac{1}{2}\\\dfrac{b}{c+a}=\dfrac{1}{2}\\\dfrac{c}{a+b}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\c+a=2b\\a+b=2c\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{2c}{c}+\dfrac{2a}{a}+\dfrac{2b}{b}=6\)