Đáp án:

$\begin{array}{l}
D = \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{19.20}}\\
 = \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \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}{{97.96}} - ... - \dfrac{1}{{3.2}} - \dfrac{1}{{2.1}}\\
 = \dfrac{1}{{99}} - \left( {\dfrac{1}{{99.98}} + \dfrac{1}{{98.97}} + \dfrac{1}{{97.96}} + ... + \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}{{96}} - \dfrac{1}{{97}} + ... + \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}}\\
E = \dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + ... + \dfrac{1}{{19.21}}\\
 = \dfrac{1}{2}.\left( {\dfrac{2}{{1.3}} + \dfrac{2}{{3.5}} + \dfrac{2}{{5.7}} + ... + \dfrac{2}{{19.21}}} \right)\\
 = \dfrac{1}{2}.\left( {1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + ... + \dfrac{1}{{19}} - \dfrac{1}{{21}}} \right)\\
 = \dfrac{1}{2}.\left( {1 - \dfrac{1}{{21}}} \right)\\
 = \dfrac{1}{2}.\dfrac{{20}}{{21}}\\
 = \dfrac{{10}}{{21}}\\
F = \dfrac{1}{{99}} - \dfrac{1}{{99.97}} - \dfrac{1}{{97.95}} - .... - \dfrac{1}{{5.3}} - \dfrac{1}{{3.1}}\\
 = \dfrac{1}{{99}} - \left( {\dfrac{1}{{99.97}} + \dfrac{1}{{97.95}} + .... + \dfrac{1}{{5.3}} + \dfrac{1}{{3.1}}} \right)\\
 = \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {\dfrac{2}{{99.97}} + \dfrac{2}{{97.95}} + ... + \dfrac{2}{{5.3}} + \dfrac{2}{{3.1}}} \right)\\
 = \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {\dfrac{1}{{97}} - \dfrac{1}{{99}} + \dfrac{1}{{95}} - \dfrac{1}{{97}} + ... + \dfrac{1}{3} - \dfrac{1}{5} + 1 - \dfrac{1}{3}} \right)\\
 = \dfrac{1}{{99}} - \dfrac{1}{2}.\left( {1 - \dfrac{1}{{99}}} \right)\\
 = \dfrac{1}{{99}} - \dfrac{1}{2}.\dfrac{{98}}{{99}}\\
 = \dfrac{1}{{99}} - \dfrac{{49}}{{99}}\\
 = \dfrac{{ - 48}}{{99}}
\end{array}$