Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x = \dfrac{5}{{12}}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{{121}}{{10}}\\
x =  - \dfrac{{137}}{{10}}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{7}{6}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = 16\\
x =  - \dfrac{4}{3}
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
e)\left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.\\
f)\left[ \begin{array}{l}
x =  - \dfrac{1}{6}\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
\left| {x - \dfrac{2}{3}} \right| = \dfrac{1}{4}\\
 \to \left[ \begin{array}{l}
x - \dfrac{2}{3} = \dfrac{1}{4}\\
x - \dfrac{2}{3} =  - \dfrac{1}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x = \dfrac{5}{{12}}
\end{array} \right.\\
b)\left| {x + \dfrac{4}{5}} \right| = \dfrac{{129}}{{10}}\\
 \to \left[ \begin{array}{l}
x + \dfrac{4}{5} = \dfrac{{129}}{{10}}\\
x + \dfrac{4}{5} =  - \dfrac{{129}}{{10}}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{121}}{{10}}\\
x =  - \dfrac{{137}}{{10}}
\end{array} \right.\\
c)\left| {3 - 2x} \right| = \left| {x - \dfrac{1}{2}} \right|\\
 \to 9 - 12x + 4{x^2} = {x^2} - x + \dfrac{1}{4}\\
 \to 3{x^2} - 11x + \dfrac{{35}}{4} = 0\\
 \to 12{x^2} - 44x + 35 = 0\\
 \to 12{x^2} - 30x - 14x + 35 = 0\\
 \to 6x\left( {2x - 5} \right) - 7\left( {2x - 5} \right) = 0\\
 \to \left( {2x - 5} \right)\left( {6x - 7} \right) = 0\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{7}{6}
\end{array} \right.\\
d)\left| {\dfrac{1}{2}x + 5} \right| = \left| {x - 3} \right|\\
 \to \dfrac{1}{4}{x^2} + 5x + 25 = {x^2} - 6x + 9\\
 \to {x^2} + 20x + 100 = 4{x^2} - 24x + 36\\
 \to 3{x^2} - 44x - 64 = 0\\
 \to 3{x^2} - 48x + 4x - 64 = 0\\
 \to 3x\left( {x - 16} \right) + 4\left( {x - 16} \right) = 0\\
 \to \left( {x - 16} \right)\left( {3x + 4} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 16\\
x =  - \dfrac{4}{3}
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
e)\left| {2x - 7} \right| = x - 2\\
 \to \left[ \begin{array}{l}
2x - 7 = x - 2\\
2x - 7 =  - x + 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
3x = 9
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.\\
f)\left| {\dfrac{1}{2} - 2x} \right| = x + 1\\
 \to \left[ \begin{array}{l}
\dfrac{1}{2} - 2x = x + 1\\
\dfrac{1}{2} - 2x =  - x - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3x =  - \dfrac{1}{2}\\
x = \dfrac{3}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{6}\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)