Đáp án:

\(\begin{array}{l}
a)x = \dfrac{1}{6}\\
c)x = \dfrac{{31}}{5}\\
e)x = 1\\
g)\left[ \begin{array}{l}
x = \dfrac{7}{6}\\
x =  - \dfrac{{11}}{6}
\end{array} \right.\\
b)x = \dfrac{{59}}{{60}}\\
d)x \in \emptyset \\
f)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
h)\left[ \begin{array}{l}
x = \dfrac{7}{6}\\
x =  - \dfrac{1}{6}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)x =  - \dfrac{2}{3} + \dfrac{5}{6}\\
 \to x = \dfrac{1}{6}\\
c)\dfrac{x}{6} =  - \dfrac{7}{{15}} + \dfrac{3}{2}\\
 \to \dfrac{x}{6} = \dfrac{{31}}{{30}}\\
 \to 5x = 31\\
 \to x = \dfrac{{31}}{5}\\
e)\dfrac{{x + 1}}{8} = \dfrac{1}{4}\\
 \to x + 1 = 2\\
 \to x = 1\\
g){\left( {x + \dfrac{1}{3}} \right)^2} = \dfrac{9}{4}\\
 \to \left| {x + \dfrac{1}{3}} \right| = \dfrac{3}{2}\\
 \to \left[ \begin{array}{l}
x + \dfrac{1}{3} = \dfrac{3}{2}\\
x + \dfrac{1}{3} =  - \dfrac{3}{2}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{7}{6}\\
x =  - \dfrac{{11}}{6}
\end{array} \right.\\
b) - x - \dfrac{1}{6} =  - \dfrac{{23}}{{20}}\\
 \to x =  - \dfrac{1}{6} + \dfrac{{23}}{{20}}\\
 \to x = \dfrac{{59}}{{60}}\\
d)\left| {x - 5} \right| =  - \dfrac{{17}}{{12}}\left( {KTM} \right)\\
Do:\left| {x - 5} \right| \ge 0\forall x\\
 \to x \in \emptyset \\
f)\left| {x - \dfrac{1}{2}} \right| = \dfrac{5}{{20}} + \dfrac{3}{4}\\
 \to \left| {x - \dfrac{1}{2}} \right| = 1\\
 \to \left[ \begin{array}{l}
x - \dfrac{1}{2} = 1\\
x - \dfrac{1}{2} =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
h){\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{8}{9} - \dfrac{4}{9}\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{4}{9}\\
 \to \left| {x - \dfrac{1}{2}} \right| = \dfrac{2}{3}\\
 \to \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{2}{3}\\
x - \dfrac{1}{2} =  - \dfrac{2}{3}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{7}{6}\\
x =  - \dfrac{1}{6}
\end{array} \right.
\end{array}\)