Đáp án:

$\begin{array}{l}
e)\left| {x + 5} \right| = \dfrac{1}{7} - \left| {\dfrac{4}{3} - \dfrac{1}{6}} \right|\\
 \Leftrightarrow \left| {x + 5} \right| = \dfrac{1}{7} - \dfrac{7}{6}\\
 \Leftrightarrow \left| {x + 5} \right| = \dfrac{{ - 43}}{{42}}\\
 \Leftrightarrow x \in \emptyset \\
Vậy\,x \in \emptyset \\
f)\left| {x + \dfrac{2}{3}} \right| = \dfrac{1}{2} - \left( {\dfrac{1}{4} - \dfrac{2}{3}} \right)\\
 \Leftrightarrow \left| {x + \dfrac{2}{3}} \right| = \dfrac{1}{2} + \dfrac{5}{{12}}\\
 \Leftrightarrow \left| {x + \dfrac{2}{3}} \right| = \dfrac{{11}}{{12}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{2}{3} = \dfrac{{11}}{{12}}\\
x + \dfrac{2}{3} = \dfrac{{ - 11}}{{12}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = \dfrac{{ - 19}}{{12}}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{4},x = \dfrac{{ - 19}}{{12}}\\
g)\left| {x - \dfrac{1}{5}} \right| = \left| {\dfrac{{ - 5}}{2}} \right| - \left( {\dfrac{1}{4} - \dfrac{2}{3}} \right)\\
 \Leftrightarrow \left| {x - \dfrac{1}{5}} \right| = \dfrac{5}{2} + \dfrac{5}{{12}}\\
 \Leftrightarrow \left| {x - \dfrac{1}{5}} \right| = \dfrac{{35}}{{12}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{5} = \dfrac{{35}}{{12}}\\
x - \dfrac{1}{5} = \dfrac{{ - 35}}{{12}}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{187}}{{60}}\\
x = \dfrac{{ - 163}}{{60}}
\end{array} \right.\\
Vậy\,x = \dfrac{{187}}{{60}},x = \dfrac{{ - 163}}{{60}}\\
h)\left| {x - \dfrac{5}{2}} \right| = \dfrac{4}{3} - \left( {\dfrac{2}{3} - \dfrac{1}{2}} \right)\\
 \Leftrightarrow \left| {x - \dfrac{5}{2}} \right| = \dfrac{4}{3} - \dfrac{1}{6}\\
 \Leftrightarrow \left| {x - \dfrac{5}{2}} \right| = \dfrac{7}{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{5}{2} = \dfrac{7}{6}\\
x - \dfrac{5}{2} = \dfrac{{ - 7}}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{11}}{3},x = \dfrac{4}{3}\\
i)\left| {x + \dfrac{5}{6}} \right| = \left| {\dfrac{1}{5} - \dfrac{2}{3}} \right| + \dfrac{{ - 3}}{4}\\
 \Leftrightarrow \left| {x + \dfrac{5}{6}} \right| = \dfrac{7}{{15}} - \dfrac{3}{4}\\
 \Leftrightarrow \left| {x + \dfrac{5}{6}} \right| = \dfrac{{ - 17}}{{60}}\\
 \Leftrightarrow x \in \emptyset \\
Vậy\,x \in \emptyset \\
k)\left| {x - \dfrac{1}{5}} \right| + \dfrac{1}{3} = \dfrac{1}{4} - \left| {\dfrac{{ - 3}}{2}} \right|\\
 \Leftrightarrow \left| {x - \dfrac{1}{5}} \right| + \dfrac{1}{3} = \dfrac{1}{4} - \dfrac{3}{2}\\
 \Leftrightarrow \left| {x - \dfrac{1}{5}} \right| = \dfrac{{ - 19}}{{12}}\\
 \Leftrightarrow x \in \emptyset \\
Vậy\,x \in \emptyset 
\end{array}$