Đáp án: B>A
Giải thích các bước giải:
$\begin{array}{l}
A = \frac{2}{{60.63}} + \frac{2}{{63.66}} + ... + \frac{2}{{117.120}} + \frac{2}{{2011}}\\
= \frac{2}{3}\left( {\frac{3}{{60.63}} + \frac{3}{{63.66}} + ... + \frac{3}{{117.120}}} \right) + \frac{2}{{2011}}\\
= \frac{2}{3}.\left( {\frac{1}{{60}} - \frac{1}{{63}} + \frac{1}{{63}} - \frac{1}{{66}} + ... + \frac{1}{{117}} - \frac{1}{{120}}} \right) + \frac{2}{{2011}}\\
= \frac{2}{3}\left( {\frac{1}{{60}} - \frac{1}{{120}}} \right) + \frac{2}{{2011}}\\
= \frac{2}{3}.\frac{1}{{120}} + \frac{2}{{2011}}\\
= \frac{1}{{180}} + \frac{2}{{2011}}\\
B = \frac{5}{{40.44}} + \frac{5}{{44.48}} + ... + \frac{5}{{76.80}} + \frac{5}{{20.10}}\\
= \frac{5}{4}\left( {\frac{1}{{40}} - \frac{1}{{44}} + \frac{1}{{44}} - \frac{1}{{48}} + ... + \frac{1}{{76}} - \frac{1}{{80}}} \right) + \frac{5}{{2010}}\\
= \frac{5}{4}\left( {\frac{1}{{40}} - \frac{1}{{80}}} \right) + \frac{5}{{2010}}\\
= \frac{5}{4}.\frac{1}{{80}} + \frac{5}{{2010}}\\
= \frac{5}{{320}} + \frac{5}{{2010}} > \frac{1}{{180}} + \frac{2}{{2011}}\\
\Rightarrow B > A
\end{array}$
