Đáp án:

 $b) x=\dfrac{-9}{8}\\
c) x=\dfrac{5}{6}\\
d) x=\dfrac{19}{18}\\
e)
 {\left[\begin{aligned}x=\dfrac{4}{6}=\dfrac{2}{3}\\x=\dfrac{-10}{6}=\dfrac{-5}{3}\end{aligned}\right.}\\
f)
{\left[\begin{aligned}x =\dfrac{13}{40}\\x =\dfrac{27}{40}\end{aligned}\right.}\\
g)
 x=\dfrac{1}{9}\\
h)
 x=\dfrac{1}{8}\\
e)
 x=\dfrac{-7}{25}$

Giải thích các bước giải:

 $b) 0,5-\dfrac{1}{3}x=\dfrac{7}{8}\\
\Leftrightarrow \dfrac{1}{3}x=\dfrac{1}{2}-\dfrac{7}{8}\\
\Leftrightarrow \dfrac{1}{3}x=\dfrac{4}{8}-\dfrac{7}{8}\\
\Leftrightarrow \dfrac{1}{3}x=\dfrac{-3}{8}\\
\Leftrightarrow x=\dfrac{-3}{8}.3=\dfrac{-9}{8}\\
c) \dfrac{1}{2}x+\dfrac{1}{3}=\left | -\dfrac{3}{4} \right |\\
\Leftrightarrow \dfrac{1}{2}x=\dfrac{3}{4}-\dfrac{1}{3}\\
\Leftrightarrow \dfrac{1}{2}x=\dfrac{9}{12}-\dfrac{4}{12}\\
\Leftrightarrow \dfrac{1}{2}x=\dfrac{5}{12}\\
\Leftrightarrow x=\dfrac{5}{12}.2=\dfrac{5}{6}\\
d) \dfrac{3}{4}.\left ( x+\dfrac{1}{2} \right )-\dfrac{1}{3}=\dfrac{5}{6}\\
\Leftrightarrow \dfrac{3}{4}.\left ( x+\dfrac{1}{2} \right )=\dfrac{5}{6}+\dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{4}.\left ( x+\dfrac{1}{2} \right )=\dfrac{5}{6}+\dfrac{2}{6}\\
\Leftrightarrow  \dfrac{3}{4}.\left ( x+\dfrac{1}{2} \right )=\dfrac{7}{6}\\
\Leftrightarrow x+\dfrac{1}{2}=\dfrac{7}{6}.\dfrac{4}{3}\\
\Leftrightarrow x=\dfrac{14}{9}-\dfrac{1}{2}\\
\Leftrightarrow x=\dfrac{28}{18}-\dfrac{9}{18}\\
\Leftrightarrow x=\dfrac{19}{18}\\
e)
\left | x+\dfrac{1}{2} \right |-\dfrac{1}{2}=\dfrac{2}{3}\\
\Leftrightarrow \left | x+\dfrac{1}{2} \right |=\dfrac{2}{3}+\dfrac{1}{2}\\
\Leftrightarrow \left | x+\dfrac{1}{2} \right |=\dfrac{4}{6}+\dfrac{3}{6}\\
\Leftrightarrow \left | x+\dfrac{1}{2} \right |=\dfrac{7}{6}\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{6}-\dfrac{1}{2}\\x=-\dfrac{7}{6}-\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{6}-\dfrac{3}{6}\\x=-\dfrac{7}{6}-\dfrac{3}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{4}{6}=\dfrac{2}{3}\\x=\dfrac{-10}{6}=\dfrac{-5}{3}\end{aligned}\right.}\\
f)
2\left | \dfrac{1}{2}-x \right |+\dfrac{2}{5}=\dfrac{3}{4}\\
\Leftrightarrow 2\left | \dfrac{1}{2}-x \right |=\dfrac{3}{4}-\dfrac{2}{5}\\
\Leftrightarrow 2\left | \dfrac{1}{2}-x \right |=\dfrac{15}{20}-\dfrac{8}{20}\\
\Leftrightarrow 2\left | \dfrac{1}{2}-x \right |=\dfrac{7}{20}\\
\Leftrightarrow \left | \dfrac{1}{2}-x \right |=\dfrac{7}{40}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{2}-x =\dfrac{7}{40}\\\dfrac{1}{2}-x =-\dfrac{7}{40}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{1}{2}-\dfrac{7}{40}\\x =\dfrac{1}{2}+\dfrac{7}{40}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{20}{40}-\dfrac{7}{40}\\x =\dfrac{20}{40}+\dfrac{7}{40}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =\dfrac{13}{40}\\x =\dfrac{27}{40}\end{aligned}\right.}\\
g)
2x-\dfrac{1}{2}=\dfrac{1}{2}x-\dfrac{1}{3}\\
\Leftrightarrow 2x-\dfrac{1}{2}x=\dfrac{-1}{3}+\dfrac{1}{2}\\
\Leftrightarrow \left (2-\dfrac{1}{2}  \right )x=\dfrac{-2}{6}+\dfrac{3}{6}\\
\Leftrightarrow  \left (\dfrac{4}{2}-\dfrac{1}{2}  \right )x=\dfrac{1}{6}\\
\Leftrightarrow \dfrac{3}{2}x=\dfrac{1}{6}\\
\Leftrightarrow x=\dfrac{1}{6}.\dfrac{2}{3}\\
\Leftrightarrow x=\dfrac{1}{9}\\
h)
\dfrac{3}{2}x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{1}{2}x\\
\Leftrightarrow \dfrac{3}{2}x+\dfrac{1}{2}x=\dfrac{3}{4}-\dfrac{1}{2}\\
\Leftrightarrow \left (\dfrac{3}{2}+\dfrac{1}{2}  \right )x=\dfrac{3}{4}-\dfrac{2}{4}\\
\Leftrightarrow \left (\dfrac{3}{2}+\dfrac{1}{2}  \right )x=\dfrac{1}{4}\\
\Leftrightarrow \dfrac{4}{2}x=\dfrac{1}{4}\\
\Leftrightarrow 2x=\dfrac{1}{4}\\
\Leftrightarrow x=\dfrac{1}{8}\\
e)
\dfrac{3}{5}-2x=3x+2\\
\Leftrightarrow 3x+2x=\dfrac{3}{5}-2\\
\Leftrightarrow 5x=\dfrac{3}{5}-\dfrac{10}{5}\\
\Leftrightarrow 5x=\dfrac{-7}{5}\\
\Leftrightarrow x=\dfrac{-7}{25}$