Đáp án:
$\begin{array}{l}
a)\left| {\dfrac{1}{2}x + \dfrac{2}{3}} \right| = \left| {\dfrac{3}{4}x + \dfrac{5}{6}} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x + \dfrac{2}{3} = \dfrac{3}{4}x + \dfrac{5}{6}\\
\dfrac{1}{2}x + \dfrac{2}{3} = - \dfrac{3}{4}x - \dfrac{5}{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{4}x - \dfrac{1}{2}x = \dfrac{2}{3} - \dfrac{5}{6}\\
\dfrac{1}{2}x + \dfrac{3}{4}x = - \dfrac{5}{6} - \dfrac{2}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{4}x = \dfrac{{ - 1}}{6}\\
\dfrac{5}{4}x = \dfrac{{ - 3}}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{2}{3}\\
x = - \dfrac{6}{5}
\end{array} \right.\\
Vậy\,x = - \dfrac{2}{3};x = - \dfrac{6}{5}
\end{array}$
$\begin{array}{l}
b)\left| {\dfrac{1}{2}x + \dfrac{5}{6}} \right| - \left| {\dfrac{7}{8}x - \dfrac{8}{9}} \right| = 0\\
\Leftrightarrow \left| {\dfrac{1}{2}x + \dfrac{5}{6}} \right| = \left| {\dfrac{7}{8}x - \dfrac{8}{9}} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2}x + \dfrac{5}{6} = \dfrac{7}{8}x - \dfrac{8}{9}\\
\dfrac{1}{2}x + \dfrac{5}{6} = \dfrac{8}{9} - \dfrac{7}{8}x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{7}{8}x - \dfrac{1}{2}x = \dfrac{5}{6} + \dfrac{8}{9}\\
\dfrac{1}{2}x + \dfrac{7}{8}x = \dfrac{8}{9} - \dfrac{5}{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{8}x = \dfrac{{31}}{{18}}\\
\dfrac{{11}}{8}x = \dfrac{1}{{18}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{124}}{{27}}\\
x = \dfrac{4}{{99}}
\end{array} \right.\\
Vậy\,x = \dfrac{{124}}{{27}};x = \dfrac{4}{{99}}
\end{array}$
