Question

[Mai Sương Nguyễn]

Cho a+b+c=2007 và  \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+d}=\frac{1}{90}\)

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

Answer 1

[tth_new]

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

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

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\Rightarrow S=\frac{223}{10}-3=\frac{193}{10}\)

Answer 2

[Napkin ( Fire Smoke Team )]

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

\(=>S+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\)

\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)