Ta có:

        $\frac{1}{a+b}$ + $\frac{1}{a+c}$ +$\frac{1}{b+c}$ =$\frac{1}{100}$ 

=>100.($\frac{1}{a+b}$ + $\frac{1}{a+c}$ + $\frac{1}{b+c}$)=1

=>$\frac{100}{a+b}$ + $\frac{100}{a+c}$ + $\frac{100}{b+c}$=1

=>$\frac{a+b+c}{a+b}$ + $\frac{a+b+c}{a+c}$ + $\frac{a+b+c}{b+c}$=1

=>$\frac{a+b}{a+b}$+$\frac{c}{a+b}$ + $\frac{a+c}{a+c}$+$\frac{b}{a+c}$+ $\frac{b+c}{b+c}$+ +$\frac{a}{b+c}$=1

=>1+$\frac{c}{b+c}$ +1+$\frac{b}{a+c}$+1+$\frac{a}{b+c}$=1

=>3+$\frac{c}{b+c}$+$\frac{b}{a+c}$+$\frac{a}{b+c}$=1

=>$\frac{c}{b+c}$+$\frac{b}{a+c}$+$\frac{a}{b+c}$=1-3

=>$\frac{c}{b+c}$+$\frac{b}{a+c}$+$\frac{a}{b+c}$=-2

Vậy S=-2