Đáp án:
$\begin{array}{l}
1)x + \dfrac{1}{2} = {2^5}:{2^3}\\
\Leftrightarrow x + \dfrac{1}{2} = {2^2}\\
\Leftrightarrow x = 4 - \dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
2)\left( {5x - 1} \right)\left( {2x - \dfrac{1}{3}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
5x - 1 = 0\\
2x - \dfrac{1}{3} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x = \dfrac{1}{6}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{5};x = \dfrac{1}{6}\\
3)\left| {x + 5} \right| - 6 = 9\\
\Leftrightarrow \left| {x + 5} \right| = 15\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 15\\
x + 5 = - 15
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x = - 20
\end{array} \right.\\
Vậy\,x = 10;x = - 20\\
4){\left( {x - \dfrac{1}{2}} \right)^3} = \dfrac{1}{{27}}\\
\Leftrightarrow x - \dfrac{1}{2} = \dfrac{1}{3}\\
\Leftrightarrow x = \dfrac{1}{3} + \dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{5}{6}\\
Vậy\,x = \dfrac{5}{6}\\
5){\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{4}{{25}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{2}{5}\\
x + \dfrac{1}{2} = - \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{5} - \dfrac{1}{2} = \dfrac{{ - 1}}{{10}}\\
x = \dfrac{{ - 2}}{5} - \dfrac{1}{2} = \dfrac{{ - 9}}{{10}}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 1}}{{10}};x = \dfrac{{ - 9}}{{10}}\\
6){2^{x - 1}} = 16\\
\Leftrightarrow {2^{x - 1}} = {2^4}\\
\Leftrightarrow x - 1 = 4\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
7){\left( {x - 1} \right)^2} = 25\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 5\\
x - 1 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 4
\end{array} \right.\\
Vậy\,x = 6;x = - 4\\
8){\left( {x - 1} \right)^{x + 2}} = {\left( {x - 1} \right)^{x + 6}}\\
\Leftrightarrow {\left( {x - 1} \right)^{x + 2}}.\left[ {1 - {{\left( {x - 1} \right)}^4}} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
{\left( {x - 1} \right)^4} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0
\end{array} \right.\\
Vậy\,x = 0;x = 1;x = 2
\end{array}$
