Đáp án:
$\begin{array}{l}
A = \dfrac{{13}}{{25}} + \dfrac{9}{{10}} - \dfrac{{11}}{{15}} + \dfrac{{13}}{{21}} - \dfrac{{15}}{{28}} + \dfrac{{17}}{{26}} - ... + \dfrac{{197}}{{4851}} - \dfrac{{199}}{{4950}}\\
= \dfrac{{13}}{{25}} + \left( {\dfrac{9}{{10}} - \dfrac{{11}}{{15}}} \right) + \left( {\dfrac{{13}}{{21}} - \dfrac{{15}}{{28}}} \right) + ... + \left( {\dfrac{{197}}{{4851}} - \dfrac{{199}}{{4950}}} \right)\\
= \dfrac{{13}}{{25}} + \dfrac{1}{6} + \dfrac{1}{{12}} + ... + \dfrac{1}{{2450}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{49.50}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{49}} - \dfrac{1}{{50}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{2} - \dfrac{1}{{50}}\\
= 1 > \dfrac{9}{{10}}\\
Vậy\,A > \dfrac{9}{{10}}
\end{array}$
