Giải:
Ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{90}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{a+b+c}{90}\)
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{90}\)
\(\Leftrightarrow\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1=\dfrac{2007}{90}\)
\(\Leftrightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{2007}{90}-3\)
\(\Leftrightarrow S=19,3\)
Vậy ...