Đáp án:
$\begin{array}{l}
b)x:\left( { - 2\dfrac{1}{{15}}} \right) + 3\dfrac{1}{2} = - \dfrac{3}{4}\\
\Leftrightarrow x:\dfrac{{ - 31}}{{15}} = - \dfrac{3}{4} - \dfrac{7}{2}\\
\Leftrightarrow x:\dfrac{{ - 31}}{{15}} = \dfrac{{ - 17}}{4}\\
\Leftrightarrow x = \dfrac{{ - 17}}{4}.\dfrac{{ - 31}}{{15}}\\
\Leftrightarrow x = \dfrac{{527}}{{60}}\\
\text{Vậy}\,x = \dfrac{{527}}{{60}}\\
b)\dfrac{5}{8} - x:1\dfrac{5}{{11}} = 2\\
\Leftrightarrow x:\dfrac{{16}}{{11}} = \dfrac{5}{8} - 2\\
\Leftrightarrow x:\dfrac{{16}}{{11}} = \dfrac{{ - 11}}{8}\\
\Leftrightarrow x = \dfrac{{ - 11}}{8}.\dfrac{{16}}{{11}}\\
\Leftrightarrow x = - 2\\
\text{Vậy}\,x = - 2\\
c)\left| {x - \dfrac{3}{4}} \right| - \dfrac{1}{4} = 0\\
\Leftrightarrow \left| {x - \dfrac{3}{4}} \right| = \dfrac{1}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{3}{4} = \dfrac{1}{4}\\
x - \dfrac{3}{4} = - \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4} + \dfrac{3}{4} = 1\\
x = - \dfrac{1}{4} + \dfrac{3}{4} = \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{1}{2};x = 1\\
d)\dfrac{3}{4}:2\dfrac{4}{9} - \left| { - 3x + 2\dfrac{2}{3}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \dfrac{3}{4}:\dfrac{{22}}{9} - \left| { - 3x + \dfrac{8}{3}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left| { - 3x + \dfrac{8}{3}} \right| = \dfrac{3}{4}.\dfrac{9}{{22}} - \dfrac{3}{4}\\
\Leftrightarrow \left| { - 3x + \dfrac{8}{3}} \right| = \dfrac{{ - 39}}{{88}}\left( {ktm} \right)\\
Do:\left| { - 3x + \dfrac{8}{3}} \right| \ge 0\\
\text{Vậy pt vô nghiệm}\\
e)\left| {x - \dfrac{1}{3}} \right| = \left| {2 - 3x} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{3} = 2 - 3x\\
x - \dfrac{1}{3} = 3x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{{12}}\\
x = \dfrac{5}{6}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{7}{{12}};x = \dfrac{5}{6}\\
f)\left( {3\dfrac{5}{7}x - 1\dfrac{5}{7}x} \right) - \dfrac{1}{3} = \dfrac{2}{3}\\
\Leftrightarrow \left( {3\dfrac{5}{7} - 1\dfrac{5}{7}} \right)x = \dfrac{2}{3} + \dfrac{1}{3}\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \dfrac{1}{2}\\
\text{Vậy}\,x = \dfrac{1}{2}
\end{array}$
