Đáp án:
$\begin{array}{l}
a)\dfrac{{{8^2}{{.8}^3}}}{{{2^8}{{.2}^7}}} = \dfrac{{{2^6}{{.2}^9}}}{{{2^8}{{.2}^7}}} = \dfrac{{{2^{15}}}}{{{2^{15}}}} = 1\\
b)\dfrac{{{2^5}{{.6}^3}}}{{{8^3}{{.9}^2}}} = \dfrac{{{2^5}{{.2}^3}{{.3}^3}}}{{{2^9}{{.3}^4}}} = \dfrac{{{2^8}{{.3}^3}}}{{{2^9}{{.3}^4}}} = \dfrac{1}{{2.3}} = \dfrac{1}{6}\\
c)\dfrac{{{9^8}{{.8}^6}}}{{{{16}^4}{{.3}^{17}}}} = \dfrac{{{3^{16}}{{.2}^{18}}}}{{{2^{16}}{{.3}^{17}}}} = \dfrac{{{2^2}}}{3} = \dfrac{4}{3}\\
d)\dfrac{{{{16}^{11}}.{{\left( { - 5} \right)}^{40}}}}{{{{\left( { - 10} \right)}^{41}}}}\\
= - \dfrac{{{2^{44}}{{.5}^{40}}}}{{{2^{41}}{{.5}^{41}}}}\\
= - \dfrac{{{2^3}}}{5}\\
= \dfrac{{ - 8}}{5}\\
e)\dfrac{{{4^5}{{.9}^4} - {{2.6}^9}}}{{{2^{10}}{{.3}^8} + {6^8}.20}}\\
= \dfrac{{{2^{10}}{{.3}^8} - {{2.2}^9}{{.3}^9}}}{{{2^{10}}{{.3}^8} + {2^8}{{.3}^8}{{.2}^2}.5}}\\
= \dfrac{{{2^{10}}{{.3}^8}\left( {1 - 3} \right)}}{{{2^{10}}{{.3}^8}.\left( {1 + 5} \right)}}\\
= \dfrac{{ - 2}}{6}\\
= - \dfrac{1}{3}\\
f)\dfrac{{{2^{10}}{{.3}^{31}} + {2^{40}}{{.3}^6}}}{{{2^{11}}{{.3}^{31}} + {2^{41}}{{.3}^6}}}\\
= \dfrac{{{2^{10}}{{.3}^6}.\left( {{3^{25}} + {2^{30}}} \right)}}{{{2^{11}}{{.3}^6}.\left( {{3^{25}} + {2^{30}}} \right)}}\\
= \dfrac{1}{2}
\end{array}$
