Đặt \(S=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{c}{c+a}\)
\(\Rightarrow S=\dfrac{2010-\left(a+b\right)}{a+b}+\dfrac{2010-\left(b+c\right)}{b+c}+\dfrac{2010-\left(c+a\right)}{c+a}\)\(\Rightarrow S=\dfrac{2010}{a+b}+\dfrac{2010}{b+c}+\dfrac{2010}{c+a}-3\)
\(\Rightarrow S=2010\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(\Rightarrow S=2010.\dfrac{1}{3}-3\)
\(\Rightarrow S=670-3\)
\(\Rightarrow S=667\)
Ta có: \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\left(\dfrac{c}{a+b}+1\right)+\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)-3\)
\(=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)-3\)
\(=2010.\dfrac{1}{3}-3\)
\(=670-3\)
\(=667\)
Vậy \(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=667\)
