Đáp án:
$\begin{array}{l}
e){\left( {\dfrac{4}{5}} \right)^{128}}:{\left( {\dfrac{{64}}{{125}}} \right)^{36}}\\
= {\left( {\dfrac{4}{5}} \right)^{128}}:{\left( {\dfrac{4}{5}} \right)^{3.36}}\\
= {\left( {\dfrac{4}{5}} \right)^{128 - 3.36}}\\
= {\left( {\dfrac{4}{5}} \right)^{10}}\\
= \dfrac{{{4^{10}}}}{{{5^{10}}}} = \dfrac{{{2^{20}}}}{{{5^{10}}}}\\
g)\dfrac{{{3^5}{{.125.32}^4}.49}}{{{9^{15}}{{.50}^4}{{.21}^4}}}\\
= \dfrac{{{3^5}{{.5}^3}.{{\left( {{2^5}} \right)}^4}{{.7}^2}}}{{{{\left( {{3^2}} \right)}^{15}}.{{\left( {{{2.5}^2}} \right)}^4}.{{\left( {3.7} \right)}^4}}}\\
= \dfrac{{{3^5}{{.5}^3}{{.2}^{20}}{{.7}^2}}}{{{3^{30}}{{.2}^4}{{.5}^8}{{.3}^4}{{.7}^4}}}\\
= \dfrac{{{2^{16}}}}{{{3^{29}}{{.5}^5}{{.7}^2}}}\\
h)\dfrac{{{9^5}{{.5}^{13}}{{.64}^4}{{.7}^2}}}{{{3^{20}}{{.10}^{10}}{{.28}^4}}}\\
= \dfrac{{{3^{10}}{{.5}^{13}}.{{\left( {{2^6}} \right)}^4}{{.7}^2}}}{{{3^{20}}{{.2}^{10}}{{.5}^{10}}.{{\left( {{2^2}.7} \right)}^4}}}\\
= \dfrac{{{3^{10}}{{.2}^{24}}{{.5}^{13}}{{.7}^2}}}{{{3^{20}}{{.2}^{18}}{{.5}^{10}}{{.7}^4}}}\\
= \dfrac{{{2^6}{{.5}^3}}}{{{3^{10}}{{.7}^2}}}
\end{array}$
