Đáp án:
$\begin{array}{l}
C = \left( {\frac{4}{3} + \frac{8}{3}} \right).\left( {\frac{7}{4} - \frac{6}{4}} \right):\left( {\frac{6}{5} + \frac{{12}}{5} + \frac{1}{5}} \right)\\
= \frac{{4 + 8}}{3}.\frac{{7 - 6}}{4}:\frac{{6 + 12 + 1}}{5}\\
= \frac{{12}}{3}.\frac{1}{4}.\frac{5}{{19}}\\
= \frac{5}{{19}}\\
D = \frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + ... + \frac{1}{{31.34}}\\
\Rightarrow 3.D = \frac{3}{{1.4}} + \frac{3}{{4.7}} + \frac{3}{{7.10}} + ... + \frac{3}{{31.34}}\\
\Rightarrow 3.D = \frac{{4 - 1}}{{1.4}} + \frac{{7 - 4}}{{4.7}} + \frac{{10 - 7}}{{7.10}} + ... + \frac{{34 - 31}}{{31.34}}\\
\Rightarrow 3.D = 1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + ... + \frac{1}{{31}} - \frac{1}{{34}}\\
\Rightarrow 3.D = 1 - \frac{1}{{34}} = \frac{{33}}{{34}}\\
\Rightarrow D = \frac{{33}}{{34}}.\frac{1}{3} = \frac{{11}}{{34}}\\
E = \frac{{1.2}}{{{1^2}}}.\frac{{2.3}}{{{2^2}}}.\frac{{3.4}}{{{3^2}}}.\frac{{4.5}}{{{4^2}}}.\frac{{5.6}}{{{5^2}}}\\
= \frac{2}{1}.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}\\
= 6
\end{array}$
