Đáp án:
$\begin{array}{l}
a)A = \dfrac{6}{{2.5}} + \dfrac{6}{{5.8}} + \dfrac{6}{{8.11}} + ... + \dfrac{6}{{29.32}}\\
= 2.\left( {\dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} + ... + \dfrac{3}{{29.32}}} \right)\\
= 2.\left( {\dfrac{{5 - 2}}{{2.5}} + \dfrac{{8 - 5}}{{5.8}} + \dfrac{{11 - 8}}{{8.11}} + ... + \dfrac{{32 - 29}}{{29.32}}} \right)\\
= 2.\left( {\dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + ... + \dfrac{1}{{29}} - \dfrac{1}{{32}}} \right)\\
= 2.\left( {\dfrac{1}{2} - \dfrac{1}{{32}}} \right)\\
= 2.\dfrac{{16 - 1}}{{32}}\\
= \dfrac{{15}}{{16}}\\
b)B = \dfrac{{{{15.9}^5} - {{4.9}^6}}}{{{{9.9}^4}{{.2}^5} - {3^9}.8}}\\
= \dfrac{{{{5.3.3}^{2.5}} - {{4.3}^{2.6}}}}{{{3^2}{{.3}^{2.4}}{{.2}^5} - {3^9}.8}}\\
= \dfrac{{{3^{11}}.\left( {5 - 4.3} \right)}}{{{3^{10}}.32 - {3^9}.8}}\\
= \dfrac{{{3^{11}}.\left( {5 - 12} \right)}}{{{3^9}.\left( {3.32 - 8} \right)}}\\
= \dfrac{{{3^2}.\left( { - 7} \right)}}{{88}}\\
= \dfrac{{ - 63}}{{88}}
\end{array}$
