Đáp án:
$\begin{array}{l}
B8)\\
a)\dfrac{{{{20}^5}{{.5}^{10}}}}{{{{100}^5}}} = \dfrac{{{2^{10}}{{.5}^5}{{.5}^{10}}}}{{{{\left( {{2^2}{{.5}^2}} \right)}^5}}} = \dfrac{{{2^{10}}{{.5}^{15}}}}{{{2^{10}}{{.5}^{10}}}} = {5^5}\\
b)\dfrac{{{{\left( {0,9} \right)}^5}}}{{{{\left( {0,3} \right)}^6}}} = {\left( {\dfrac{{0,9}}{{0,3}}} \right)^5}.\dfrac{1}{{0,3}} = {3^5}.\dfrac{{10}}{3} = {3^4}.10 = 810\\
c)\dfrac{{{2^{15}}{{.9}^4}}}{{{6^6}{{.8}^3}}} = \dfrac{{{2^{15}}{{.3}^8}}}{{{2^6}{{.3}^6}{{.2}^9}}} = \dfrac{{{2^{15}}{{.3}^8}}}{{{2^{15}}{{.3}^6}}} = {3^2} = 9\\
d)\dfrac{{{6^3} + {{3.6}^2} + {3^3}}}{{ - 13}} = \dfrac{{{2^3}{{.3}^3} + {2^2}{{.3}^3} + {3^3}}}{{ - 13}}\\
= \dfrac{{{3^3}\left( {{2^3} + {2^2} + 1} \right)}}{{ - 13}}\\
= \dfrac{{27.13}}{{ - 13}}\\
= - 27\\
e)\dfrac{{{4^6}{{.9}^5} + {6^9}.120}}{{{8^4}{{.3}^{12}} - {6^{11}}}}\\
= \dfrac{{{2^{12}}{{.3}^{10}} + {2^9}{{.3}^9}{{.2}^3}{{.5}^2}}}{{{2^{12}}{{.3}^{12}} - {2^{11}}{{.3}^{11}}}}\\
= \dfrac{{{2^{12}}{{.3}^{10}}\left( {1 + 3} \right)}}{{{2^{11}}{{.3}^{11}}.\left( {2.3 - 1} \right)}}\\
= \dfrac{{2.4}}{{3.5}}\\
= \dfrac{8}{{15}}\\
B9)\\
a){\left( {\dfrac{1}{2}} \right)^n} = \dfrac{1}{8}\\
\Rightarrow \dfrac{1}{{{2^n}}} = \dfrac{1}{{{2^3}}} \Rightarrow n = 3\\
b){\left( { - \dfrac{3}{4}} \right)^n} = \dfrac{{81}}{{256}}\\
\Rightarrow {\left( { - \dfrac{3}{4}} \right)^n} = {\left( { - \dfrac{3}{4}} \right)^4}\\
\Rightarrow n = 4\\
c){27^n}:{3^n} = 9\\
\Rightarrow {9^n} = 9\\
\Rightarrow n = 1\\
d)\dfrac{{25}}{{{5^n}}} = 5\\
\Rightarrow {5^{2 - n}} = 5\\
\Rightarrow 2 - n = 1\\
\Rightarrow n = 1\\
e)\dfrac{{81}}{{{{\left( { - 3} \right)}^n}}} = - 243\\
\Rightarrow {\left( { - 3} \right)^{4 - n}} = {\left( { - 3} \right)^5}\\
\Rightarrow 4 - n = 5\\
\Rightarrow n = - 1\\
f)\dfrac{1}{2}{.2^n} + {4.2^n} = {9.2^5}\\
\Rightarrow {2^{n - 1}} + {2^{n - 1}}.2.4 = {9.2^5}\\
\Rightarrow {9.2^{n - 1}} = {9.2^5}\\
\Rightarrow n - 1 = 5\\
\Rightarrow n = 6\\
g)2.16 \ge {2^n} \ge 4\\
\Rightarrow {2^5} \ge {2^n} \ge {2^2}\\
\Rightarrow 5 \ge n \ge 2\\
\Rightarrow n \in \left\{ {2;3;4;5} \right\}\\
h)32 < {2^n} < 512\\
\Rightarrow {2^5} < {2^n} < {5^9}\\
\Rightarrow 5 < n < 9\\
\Rightarrow n \in \left\{ {6;7;8} \right\}
\end{array}$
