Đáp án:
$\begin{array}{l}
A = \dfrac{2}{{50.52}} + \dfrac{2}{{51.53}} + \dfrac{2}{{52.54}} + ... + \dfrac{2}{{88.90}}\\
= \dfrac{{52 - 50}}{{50.52}} + \dfrac{{53 - 51}}{{51.53}} + \dfrac{{54 - 52}}{{52.54}} + ... + \dfrac{{90 - 88}}{{88.90}}\\
= \dfrac{1}{{50}} - \dfrac{1}{{52}} + \dfrac{1}{{51}} - \dfrac{1}{{53}} + \dfrac{1}{{52}} - \dfrac{1}{{54}} + ... + \dfrac{1}{{88}} - \dfrac{1}{{90}}\\
= \dfrac{1}{{50}} - \dfrac{1}{{90}}\\
= \dfrac{4}{{450}}\\
= \dfrac{2}{{225}}
\end{array}$
