Đáp án:
$\begin{array}{l}
c)D = \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{19.20}}\\
= \dfrac{{3 - 2}}{{2.3}} + \dfrac{{4 - 3}}{{3.4}} + \dfrac{{5 - 4}}{{4.5}} + ... + \dfrac{{20 - 19}}{{19.20}}\\
= \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{19}} - \dfrac{1}{{20}}\\
= \dfrac{1}{2} - \dfrac{1}{{20}}\\
= \dfrac{9}{{20}}\\
E = \dfrac{1}{{99}} - \dfrac{1}{{99.98}} - \dfrac{1}{{98.97}} - ... - \dfrac{1}{{3.2}} - \dfrac{1}{{2.1}}\\
= \dfrac{1}{{99}} - \left( {\dfrac{1}{{99.98}} + \dfrac{1}{{98.97}} + ... + \dfrac{1}{{3.2}} + \dfrac{1}{{2.1}}} \right)\\
= \dfrac{1}{{99}} - \left( {\dfrac{1}{{98}} - \dfrac{1}{{99}} + \dfrac{1}{{97}} - \dfrac{1}{{98}} + ... + \dfrac{1}{2} - \dfrac{1}{3} + 1 - \dfrac{1}{2}} \right)\\
= \dfrac{1}{{99}} - \left( {1 - \dfrac{1}{{99}}} \right)\\
= \dfrac{1}{{99}} - 1 + \dfrac{1}{{99}}\\
= \dfrac{2}{{99}} - 1 = \dfrac{{ - 97}}{{99}}\\
P = \dfrac{2}{{2.4}} + \dfrac{2}{{4.6}} + \dfrac{2}{{6.8}} + .. + \dfrac{2}{{2008.2010}}\\
= \dfrac{{4 - 2}}{{2.4}} + \dfrac{{6 - 4}}{{4.6}} + \dfrac{{8 - 6}}{{6.8}} + ... + \dfrac{{2010 - 2008}}{{2008.2010}}\\
= \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{6} + \dfrac{1}{6} - \dfrac{1}{8} + ... + \dfrac{1}{{2008}} - \dfrac{1}{{2010}}\\
= \dfrac{1}{2} - \dfrac{1}{{2010}}\\
= \dfrac{{1004}}{{2010}} = \dfrac{{502}}{{1005}}
\end{array}$
