Ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

=>\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

=>\(S+3=\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}\)

=>\(S+3=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=\frac{3^{2003}}{3^{200}}=3^{1803}\)

=>\(S=3^{1803}-3\)

=> \(\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)

đoạn tiếp tham khảo tại: Boul đz :D

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2015.\frac{1}{90}-3=19\frac{7}{18}\)

lay ong di qua lay ba di lai cho xin may tick

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{1}{7}\left(a+b+c\right)\) (nhân a + b +c vào mỗi vế)

\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{2009}{7}\)

Suy ra \(S=\frac{2009}{7}-3=284\)