Đáp án:
$\begin{array}{l}
a)\left( {4\frac{1}{3} + 2\frac{1}{2}} \right):\frac{{ - 11}}{{12}}\\
= \left( {\frac{{13}}{3} + \frac{5}{2}} \right).\frac{{ - 12}}{{11}}\\
= \frac{{41}}{6}.\frac{{ - 12}}{{11}}\\
= \frac{{ - 82}}{{11}}\\
b)\left( { - 1\frac{1}{5}} \right).\left( { - 1.\frac{1}{6}} \right).\left( { - 1.\frac{1}{7}} \right)...\left( { - 1\frac{1}{{2011}}} \right)\\
\frac{{ - 6}}{5}.\frac{{ - 7}}{6}.\frac{{ - 8}}{7}...\frac{{ - 2012}}{{2011}}\\
= - \frac{{2012}}{5}\\
c)\left( {1 + \frac{1}{{11}}} \right).\left( {1 + \frac{1}{{12}}} \right).\left( {1 + \frac{1}{{13}}} \right).\left( {1 + \frac{1}{{14}}} \right).\left( {1 + \frac{1}{{15}}} \right).\left( {1 + \frac{1}{{16}}} \right)\\
= \frac{{12}}{{11}}.\frac{{13}}{{12}}.\frac{{14}}{{13}}.\frac{{15}}{{14}}.\frac{{16}}{{15}}.\frac{{17}}{{16}}\\
= \frac{{17}}{{11}}\\
d)\frac{2}{{1.4}} + \frac{2}{{4.7}} + \frac{2}{{7.10}} + ... + \frac{2}{{91.94}} + \frac{2}{{94.97}}\\
= \frac{2}{3}.\left( {\frac{3}{{1.4}} + \frac{3}{{4.7}} + \frac{3}{{7.10}} + ... + \frac{3}{{91.94}} + \frac{3}{{97.94}}} \right)\\
= \frac{2}{3}.\left( {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + ... + \frac{1}{{94}} - \frac{1}{{97}}} \right)\\
= \frac{2}{3}.\left( {1 - \frac{1}{{97}}} \right)\\
= \frac{2}{3}.\frac{{96}}{{97}}\\
= \frac{{64}}{{97}}
\end{array}$
