Question

[le duc minh vuong]

trả lời giúp

Answer 1

[Hoang Hung Quan]

Bài 1: b) Ta có: \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

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

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

\(=1+1+1+\left(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)\) \(=2010.\dfrac{1}{3}\)

\(\Rightarrow3+S=\dfrac{2010}{3}=670\Leftrightarrow S=667\)

Vậy \(S=667\)