Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=11\cdot\frac{13}{17}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{143}{17}\)
\(\Rightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{a+c}{c+a}=\frac{143}{17}\)
\(\Rightarrow1+1+1+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=\frac{143}{17}\)
\(\Rightarrow A=\frac{143}{17}-3=\frac{92}{17}\)