Đáp án:
$\begin{array}{l}
2a)\dfrac{2}{3} - \left( {\dfrac{1}{2}x + 1} \right).\dfrac{3}{7} = \dfrac{1}{4}\\
\Leftrightarrow \left( {\dfrac{1}{2}x + 1} \right).\dfrac{3}{7} = \dfrac{2}{3} - \dfrac{1}{4} = \dfrac{5}{{12}}\\
\Leftrightarrow \dfrac{1}{2}x + 1 = \dfrac{5}{{12}}.\dfrac{7}{3}\\
\Leftrightarrow \dfrac{1}{2}x + 1 = \dfrac{{35}}{{36}}\\
\Leftrightarrow \dfrac{1}{2}x = \dfrac{{35}}{{36}} - 1 = \dfrac{{ - 1}}{{36}}\\
\Leftrightarrow x = - \dfrac{1}{{36}}.2\\
\Leftrightarrow x = \dfrac{{ - 1}}{{18}}\\
Vậy\,x = - \dfrac{1}{{18}}\\
b){\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{1}{{16}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{1}{4}\\
x + \dfrac{1}{2} = - \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4} - \dfrac{1}{2} = - \dfrac{1}{4}\\
x = - \dfrac{1}{4} - \dfrac{1}{2} = - \dfrac{3}{4}
\end{array} \right.\\
Vậy\,x = - \dfrac{1}{4};x = - \dfrac{3}{4}\\
c)\dfrac{4}{3}.\left| {x - 4} \right| = \dfrac{4}{9}\\
\Leftrightarrow \left| {x - 4} \right| = \dfrac{1}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
x - 4 = \dfrac{1}{3}\\
x - 4 = - \dfrac{1}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4 + \dfrac{1}{3} = \dfrac{{13}}{3}\\
x = 4 - \dfrac{1}{3} = \dfrac{{11}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{13}}{3};x = \dfrac{{11}}{3}\\
3)a)\dfrac{2}{{x - 1}} < 0\\
\Leftrightarrow x - 1 < 0\\
\Leftrightarrow x < 1\\
Vậy\,x < 1\\
b)\dfrac{{ - 3}}{{x - 6}} > 0\\
\Leftrightarrow x - 6 < 0\\
\Leftrightarrow x < 6\\
Vậy\,x < 6
\end{array}$
