Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
32:{2^n} = 4\\
\Leftrightarrow {2^n} = 32:4\\
\Leftrightarrow {2^n} = 8\\
\Leftrightarrow {2^n} = {2^3}\\
\Leftrightarrow n = 3\\
b,\\
{\left( {\dfrac{1}{2}} \right)^{2n - 1}} = \dfrac{1}{8}\\
\Leftrightarrow \dfrac{1}{{{2^{2n - 1}}}} = \dfrac{1}{8}\\
\Leftrightarrow {2^{2n - 1}} = 8\\
\Leftrightarrow {2^{2n - 1}} = {2^3}\\
\Leftrightarrow 2n - 1 = 3\\
\Leftrightarrow 2n = 4\\
\Leftrightarrow n = 2\\
d,\\
{\left( {n + 5} \right)^3} = - 64\\
\Leftrightarrow {\left( {n + 5} \right)^3} = {\left( { - 4} \right)^3}\\
\Leftrightarrow n + 5 = - 4\\
\Leftrightarrow n = - 9\\
2,\\
A = \dfrac{{{2^{10}} + {4^{10}}}}{{{8^4} + {4^{11}}}} = \dfrac{{{2^{10}} + {{\left( {{2^2}} \right)}^{10}}}}{{{{\left( {{2^3}} \right)}^4} + {{\left( {{2^2}} \right)}^{11}}}} = \dfrac{{{2^{10}} + {2^{20}}}}{{{2^{12}} + {2^{22}}}} = \dfrac{{{2^{10}}.\left( {1 + {2^{10}}} \right)}}{{{2^{12}}.\left( {1 + {2^{10}}} \right)}} = \dfrac{1}{{{2^2}}} = \dfrac{1}{4}
\end{array}\)
\(\begin{array}{l}
c,\\
\dfrac{4}{2} = \dfrac{4}{{\dfrac{n}{2}}}\\
\Leftrightarrow 2 = \dfrac{4}{{\dfrac{n}{2}}}\\
\Leftrightarrow \dfrac{n}{2} = 4:2\\
\Leftrightarrow \dfrac{n}{2} = 2\\
\Leftrightarrow n = 4
\end{array}\)
