Đáp án:
$\begin{array}{l}
1)a){2^2}.\dfrac{{{2^4}}}{{32}}.8\\
= \dfrac{{{2^2}{{.2}^4}{{.2}^3}}}{{{2^5}}}\\
= {2^{2 + 4 + 3 - 5}}\\
= {2^4}\\
b){\left( {27.81} \right)^2}:{\left( {3.27} \right)^3}\\
= {\left( {{3^3}{{.3}^4}} \right)^2}:{\left( {{{3.3}^3}} \right)^3}\\
= {3^{7.2}}:{3^{4.3}}\\
= {3^{14 - 12}}\\
= {3^2}\\
c){\left( {\dfrac{1}{5}} \right)^2}.625.\dfrac{1}{{{5^2}}}.125\\
= \dfrac{1}{{{5^2}}}{.5^4}.\dfrac{1}{{{5^2}}}{.5^3}\\
= {5^{4 + 3 - 2 - 2}}\\
= {5^3}\\
B2)\\
a)x + \dfrac{1}{2} = \dfrac{{ - 3}}{7}\\
\Leftrightarrow x = \dfrac{{ - 3}}{7} - \dfrac{1}{2}\\
\Leftrightarrow x = - \dfrac{{13}}{{14}}\\
Vậy\,x = \dfrac{{ - 13}}{{14}}\\
b){\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{25}}{{36}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{5}{6}\\
x - \dfrac{1}{2} = - \dfrac{5}{6}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{6} + \dfrac{1}{2} = \dfrac{4}{3}\\
x = - \dfrac{5}{6} + \dfrac{1}{2} = \dfrac{{ - 1}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{4}{3};x = - \dfrac{1}{3}\\
c)3{\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^3} = 81\\
\Leftrightarrow {\left( {\dfrac{x}{2} + \dfrac{3}{4}} \right)^3} = 27\\
\Leftrightarrow \left( {\dfrac{x}{2} + \dfrac{3}{4}} \right) = 3\\
\Leftrightarrow \dfrac{x}{2} = 3 - \dfrac{3}{4} = \dfrac{9}{4}\\
\Leftrightarrow x = \dfrac{9}{2}\\
Vậy\,x = \dfrac{9}{2}
\end{array}$
