Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
a,\\
{2^{28}} = {\left( {{2^4}} \right)^7} = {16^7}\\
{3^{21}} = {\left( {{3^3}} \right)^7} = {27^7}\\
\Rightarrow {2^{28}} < {3^{21}} \Leftrightarrow - {2^{28}} > - {3^{21}}\\
b,\\
{8^{10}}{.3^{30}} = {\left( {{2^3}} \right)^{10}}{.3^{30}} = {2^{30}}{.3^{30}} = {6^{30}} = {35^{15}} > {25^{15}}\\
c,\\
{32^7} = {\left( {{2^5}} \right)^7} = {2^{35}}\\
{16^9} = {\left( {{2^4}} \right)^9} = {2^{36}}\\
\Rightarrow {32^7} < {16^9} \Leftrightarrow \frac{1}{{{{32}^7}}} > \frac{1}{{{{16}^9}}}
\end{array}\]
