Đáp án:
$\begin{array}{l}
d){\left( {2x - \dfrac{1}{3}} \right)^5}:\dfrac{1}{9} = \dfrac{{ - 1}}{{27}}\\
\Leftrightarrow {\left( {2x - \dfrac{1}{3}} \right)^5} = \left( {\dfrac{{ - 1}}{{27}}} \right).\dfrac{1}{9}\\
\Leftrightarrow {\left( {2x - \dfrac{1}{3}} \right)^5} = - \dfrac{1}{{81}}\\
\Leftrightarrow {\left( {2x - \dfrac{1}{3}} \right)^5} = {\left( { - \dfrac{1}{3}} \right)^5}\\
\Leftrightarrow \left( {2x - \dfrac{1}{3}} \right) = - \dfrac{1}{3}\\
\Leftrightarrow 2x = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
e){\left( { - 1,5} \right)^0} - {\left( {2\dfrac{1}{2}x + \dfrac{{0,12}}{6}} \right)^4} = - 15\\
\Leftrightarrow 1 - {\left( {\dfrac{5}{2}x + 0,02} \right)^4} = - 15\\
\Leftrightarrow {\left( {\dfrac{5}{2}x + 0,02} \right)^4} = 16\\
\Leftrightarrow {\left( {\dfrac{5}{2}x + \dfrac{1}{{50}}} \right)^4} = {2^4}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{5}{2}x + \dfrac{1}{{50}} = 2\\
\dfrac{5}{2}x + \dfrac{1}{{50}} = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{99}}{{125}}\\
x = - \dfrac{{101}}{{125}}
\end{array} \right.\\
Vậy\,x = \dfrac{{99}}{{125}};x = - \dfrac{{101}}{{125}}
\end{array}$
