\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{3}-\dfrac{1}{13}=\dfrac{13-3}{39}=\dfrac{10}{39}\)
\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)