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

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

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\Rightarrow S=\frac{223}{10}-3=\frac{193}{10}\)

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

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

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\)

\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{90}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2017}{90}\)

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2017}{90}\)

\(\Rightarrow A+3=\frac{2017}{90}\)

\(\Rightarrow S=\frac{2017}{90}-3=\frac{1747}{90}\)

từ giả thiết, ta có 

\(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}=\frac{1}{90}\)

Mà \(S=\frac{a}{2017-a}+\frac{b}{2017-b}+\frac{c}{2017-c}=-3+\frac{2017}{2017-a}+\frac{2017}{2017-b}+\frac{2017}{2017-c}\)

=-3+\(2017\left(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}\right)=-3+\frac{2017}{90}=\frac{1747}{90}\)

vậy ...

^_^

bạn bị sai đề rồi

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

\(\Rightarrow S=2007.\frac{1}{90}-3=\frac{2007-270}{90}\)

         Vì \(a+b+c=2016\Rightarrow a=2016-\left(b+c\right);b=2016-\left(a+c\right);c=2016-\left(a+b\right)\)

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

           \(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)

           \(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)

           \(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

           \(S=2016.\frac{1}{2016}-3\)

          \(S=-2\)

Sao đề lại là a,b,c=2007? ,a=b=c=2007 ak?