Đáp án:
$\begin{array}{l}
a)\dfrac{{{5^4}{{.20}^4}}}{{{{25}^5}{{.4}^5}}} = \dfrac{{{5^4}{{.4}^4}{{.5}^4}}}{{{{\left( {{5^2}} \right)}^5}{{.4}^5}}} = \dfrac{{{4^4}{{.5}^8}}}{{{4^5}{{.5}^{10}}}}\\
= \dfrac{1}{{{{4.5}^2}}} = \dfrac{1}{{100}}\\
a)\dfrac{{{{81}^{11}}{{.3}^{17}}}}{{{{27}^{10}}{{.9}^{15}}}} = \dfrac{{{{\left( {{3^4}} \right)}^{11}}{{.3}^{17}}}}{{{{\left( {{3^3}} \right)}^{10}}.{{\left( {{3^2}} \right)}^{15}}}} = \dfrac{{{3^{44}}{{.3}^{17}}}}{{{3^{30}}{{.3}^{30}}}}\\
= \dfrac{{{3^{61}}}}{{{3^{60}}}} = 6\\
a)\dfrac{{{3^{17}}{{.81}^{11}}}}{{{{27}^{10}}{{.9}^{15}}}} = \dfrac{{{3^{17}}.{{\left( {{3^4}} \right)}^{11}}}}{{{{\left( {{3^3}} \right)}^{10}}.{{\left( {{3^2}} \right)}^{15}}}} = \dfrac{{{3^{61}}}}{{{{60}^{60}}}} = 6\\
a)\dfrac{{{{45}^{10}}{{.5}^{20}}}}{{{{75}^{15}}}} = \dfrac{{{3^{20}}{{.5}^{10}}{{.5}^{20}}}}{{{3^{15}}{{.5}^{30}}}} = {3^5}\\
b){\left( {\dfrac{4}{7}} \right)^{19}}:{\left( {\dfrac{{ - 12}}{{35}}} \right)^{19}}\\
= {\left( {\dfrac{4}{7}:\dfrac{{ - 12}}{{35}}} \right)^{19}}\\
= {\left( {\dfrac{{ - 5}}{3}} \right)^{19}}\\
b){\left( {\dfrac{{ - 4}}{5}} \right)^{15}}:{\left( {\dfrac{{ - 4}}{5}} \right)^{13}}\\
= {\left( {\dfrac{{ - 4}}{5}} \right)^2}\\
b){\left( {\dfrac{{ - 10}}{3}} \right)^5}.{\left( {\dfrac{{ - 6}}{5}} \right)^4}\\
= \dfrac{{ - 10}}{3}.{\left( {\dfrac{{ - 10}}{3}.\dfrac{6}{5}} \right)^4}\\
= \dfrac{{ - 10}}{3}.{\left( { - 4} \right)^4}\\
= \dfrac{{ - 10}}{3}.256\\
= \dfrac{{ - 2560}}{3}\\
b){\left( {\dfrac{{ - 6}}{7}} \right)^{12}}:{\left( {\dfrac{{ - 84}}{3}} \right)^{12}}\\
= {\left( {\dfrac{{ - 6}}{7}:\dfrac{{ - 84}}{3}} \right)^{12}}\\
= {\left( {\dfrac{3}{{98}}} \right)^{12}}
\end{array}$
$\begin{array}{l}
c){2^2}{.2^3}.{\left( {\dfrac{2}{3}} \right)^{ - 2}}\\
= {2^5}.\dfrac{9}{4}\\
= {2^3}.9\\
= 72\\
c){125.5^2}.\dfrac{1}{{625}}\\
= {5^3}{.5^2}.\dfrac{1}{{{5^4}}}\\
= 5\\
c){3^2}.\dfrac{1}{{243}}{.81^2}.\dfrac{1}{{{3^3}}}\\
= {3^2}.\dfrac{1}{{{3^5}}}{3^8}.\dfrac{1}{{{3^3}}}\\
= {3^2} = 9\\
c)\left( {{{4.2}^5}} \right):\left( {{2^3}.\dfrac{1}{{16}}} \right)\\
= {2^7}:\dfrac{1}{2}\\
= {2^8}
\end{array}$
