Đáp án:

\(\begin{array}{l}
a)x = \dfrac{{16}}{7}\\
b)x =  - \dfrac{2}{7}\\
c)x = \dfrac{{295}}{{29}}\\
d)\left[ \begin{array}{l}
x = 8\\
x =  - 7
\end{array} \right.\\
e)\left[ \begin{array}{l}
x = \dfrac{4}{9}\\
x =  - \dfrac{5}{9}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{2}{3}.x = \dfrac{5}{3} - \dfrac{1}{7}\\
 \to \dfrac{2}{3}.x = \dfrac{{32}}{{21}}\\
 \to x = \dfrac{{32}}{{21}}:\dfrac{2}{3}\\
 \to x = \dfrac{{16}}{7}\\
b)\dfrac{2}{3}:x =  - 2 - \dfrac{1}{3}\\
 \to \dfrac{2}{3}:x =  - \dfrac{7}{3}\\
 \to x = \dfrac{2}{3}:\left( { - \dfrac{7}{3}} \right)\\
 \to x =  - \dfrac{2}{7}\\
c)\dfrac{1}{7}:\left( {\dfrac{x}{5} - 2} \right) = \dfrac{8}{7} + 3\\
 \to \dfrac{1}{7}:\left( {\dfrac{x}{5} - 2} \right) = \dfrac{{29}}{7}\\
 \to \dfrac{x}{5} - 2 = \dfrac{1}{7}:\dfrac{{29}}{7}\\
 \to \dfrac{x}{5} - 2 = \dfrac{1}{{29}}\\
 \to \dfrac{x}{5} = \dfrac{1}{{29}} + 2\\
 \to \dfrac{x}{5} = \dfrac{{59}}{{29}}\\
 \to x = \dfrac{{295}}{{29}}\\
d)DK:x \ne \dfrac{1}{2}\\
\dfrac{{2x - 1}}{{ - 5}} = \dfrac{{ - 45}}{{2x - 1}}\\
 \to {\left( {2x - 1} \right)^2} = 225\\
 \to \left| {2x - 1} \right| = 15\\
 \to \left[ \begin{array}{l}
2x - 1 = 15\\
2x - 1 =  - 15
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 8\\
x =  - 7
\end{array} \right.\\
e)\left| {3x + \dfrac{1}{6}} \right| = 5 - \dfrac{7}{2}\\
 \to \left| {3x + \dfrac{1}{6}} \right| = \dfrac{3}{2}\\
 \to \left[ \begin{array}{l}
3x + \dfrac{1}{6} = \dfrac{3}{2}\\
3x + \dfrac{1}{6} =  - \dfrac{3}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3x = \dfrac{4}{3}\\
3x =  - \dfrac{5}{3}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{4}{9}\\
x =  - \dfrac{5}{9}
\end{array} \right.
\end{array}\)