Đáp án:
$\begin{array}{l}
a)A = \frac{4}{{1.5}} + \frac{4}{{5.9}} + ... + \frac{4}{{2001.2005}}\\
= \frac{{5 - 1}}{{1.5}} + \frac{{9 - 5}}{{5.9}} + ... + \frac{{2005 - 2001}}{{2001.2005}}\\
= 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{9} + ... + \frac{1}{{2001}} - \frac{1}{{2005}}\\
= 1 - \frac{1}{{2005}}\\
= \frac{{2004}}{{2005}}\\
b)B = \frac{2}{{10.12}} + \frac{2}{{12.14}} + ... + \frac{2}{{998.1000}}\\
= \frac{{12 - 10}}{{10.12}} + \frac{{14 - 12}}{{12.14}} + ... + \frac{{1000 - 998}}{{998.1000}}\\
= \frac{1}{{10}} - \frac{1}{{12}} + \frac{1}{{12}} - \frac{1}{{14}} + ... + \frac{1}{{998}} - \frac{1}{{1000}}\\
= \frac{1}{{10}} - \frac{1}{{1000}}\\
= \frac{{100 - 1}}{{1000}}\\
= \frac{{99}}{{1000}}
\end{array}$
