Đáp án:
$\begin{array}{l}
a)x - \dfrac{5}{3} = 2x + \dfrac{1}{4}\\
\Rightarrow 2x - x = - \dfrac{5}{3} - \dfrac{1}{4}\\
\Rightarrow x = \dfrac{{ - 20}}{{12}} - \dfrac{3}{{12}}\\
\Rightarrow x = \dfrac{{ - 23}}{{12}}\\
b)\dfrac{{x - 1}}{{ - 2}} = \dfrac{{ - 8}}{{x - 1}}\\
\Rightarrow \left( {x - 1} \right).\left( {x - 1} \right) = \left( { - 8} \right).\left( { - 2} \right)\\
\Rightarrow {\left( {x - 1} \right)^2} = 16 = {4^2} = {\left( { - 4} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 4\\
x - 1 = - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4 + 1 = 5\\
x = - 4 + 1 = - 3
\end{array} \right.\\
Vậy\,x = 5;x = - 3\\
c)1 - {x^2} = \dfrac{{15}}{{16}}\\
\Rightarrow {x^2} = 1 - \dfrac{{15}}{{16}}\\
\Rightarrow {x^2} = \dfrac{1}{{16}} = {\left( {\dfrac{1}{4}} \right)^2} = {\left( { - \dfrac{1}{4}} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = - \dfrac{1}{4}
\end{array} \right.\\
d) - 2{x^3} + 3 = \dfrac{{13}}{4}\\
\Rightarrow 2{x^3} = 3 - \dfrac{{13}}{4}\\
\Rightarrow 2{x^3} = - \dfrac{1}{4}\\
\Rightarrow {x^3} = - \dfrac{1}{8} = {\left( { - \dfrac{1}{2}} \right)^3}\\
\Rightarrow x = - \dfrac{1}{2}\\
e){\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{16}}{{25}}\\
\Rightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{4}{5}\\
x - \dfrac{1}{2} = - \dfrac{4}{5}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{4}{5} + \dfrac{1}{2} = \dfrac{{13}}{{10}}\\
x = - \dfrac{4}{5} + \dfrac{1}{2} = \dfrac{{ - 3}}{{10}}
\end{array} \right.
\end{array}$
