Đáp án:
$\begin{array}{l}
a)B = \frac{{{{10}^{2015}}}}{{{{10}^{2015}} - 5}} > 1\\
\Rightarrow \frac{{{{10}^{2015}}}}{{{{10}^{2015}} - 5}} > \frac{{{{10}^{2015}} + 2}}{{{{10}^{2015}} - 5 + 2}}\\
\Rightarrow B > \frac{{{{10}^{2015}} + 2}}{{{{10}^{1015}} - 3}}\\
\Rightarrow B > A\\
b)A = {\left( {\frac{1}{{32}}} \right)^5} = \frac{1}{{{{\left( {{2^5}} \right)}^5}}} = \frac{1}{{{2^{25}}}}\\
B = {\left( {\frac{1}{{16}}} \right)^7} = \frac{1}{{{{\left( {{2^4}} \right)}^7}}} = \frac{1}{{{2^{28}}}}\\
Vi:{2^{25}} < {2^{28}}\\
\Rightarrow \frac{1}{{{2^{25}}}} > \frac{1}{{{2^{28}}}}\\
\Rightarrow A > B
\end{array}$
