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 \(S=667\)
Đúng 0
Bình luận (0)
