Đáp án:
$\begin{array}{l}
a)2\dfrac{1}{4} + 3x = - 0,75\\
\Leftrightarrow \dfrac{9}{4} + 3x = - \dfrac{3}{4}\\
\Leftrightarrow 3x = \dfrac{9}{4} + \dfrac{3}{4} = 3\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
c)\dfrac{1}{4} - \left| {x + \dfrac{1}{2}} \right| = - \dfrac{5}{6}\\
\Leftrightarrow \left| {x + \dfrac{1}{2}} \right| = \dfrac{1}{4} + \dfrac{5}{6}\\
\Leftrightarrow \left| {x + \dfrac{1}{2}} \right| = \dfrac{{13}}{{12}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{{13}}{{12}}\\
x + \dfrac{1}{2} = - \dfrac{{13}}{{12}}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{13}}{{12}} - \dfrac{1}{2} = \dfrac{7}{{12}}\\
x = - \dfrac{{13}}{{12}} - \dfrac{1}{2} = \dfrac{{ - 19}}{{12}}
\end{array} \right.\\
Vậy\,x = \dfrac{7}{{12}};x = - \dfrac{{19}}{{12}}\\
d)\dfrac{1}{2} - {\left( {2x - \dfrac{1}{3}} \right)^2} = \dfrac{7}{{18}}\\
\Leftrightarrow {\left( {2x - \dfrac{1}{3}} \right)^2} = \dfrac{1}{2} - \dfrac{7}{{18}}\\
\Leftrightarrow {\left( {2x - \dfrac{1}{3}} \right)^2} = \dfrac{2}{{18}} = \dfrac{1}{9}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = \dfrac{1}{3}\\
2x - \dfrac{1}{3} = - \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{2}{3}\\
2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 0
\end{array} \right.\\
Vậy\,x = \dfrac{1}{3};x = 0
\end{array}$
