Question

Cho \(a+b+c=2009\)

và \(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+a}=\frac{1}{7}\)

Tính \(S=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)

Answer 1

Ta có :

\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)

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

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

\(=2009.\frac{1}{7}=287\)

\(\Rightarrow S=287-3=284\)