Đáp án:

$\begin{array}{l}
a) - \dfrac{1}{{10}} + \dfrac{2}{5}x + \dfrac{7}{{20}} = \dfrac{1}{{10}}\\
 \Leftrightarrow \dfrac{2}{5}x = \dfrac{1}{{10}} + \dfrac{1}{{10}} - \dfrac{7}{{20}}\\
 \Leftrightarrow \dfrac{2}{5}x = \dfrac{{ - 3}}{{20}}\\
 \Leftrightarrow x =  - \dfrac{3}{{20}}.\dfrac{5}{2} = \dfrac{{ - 3}}{8}\\
Vậy\,x = \dfrac{{ - 3}}{8}\\
b)\dfrac{1}{3} + \dfrac{1}{2}:x = \dfrac{{ - 1}}{5}\\
 \Leftrightarrow \dfrac{1}{2}:x = \dfrac{{ - 1}}{5} - \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{1}{2}:x = \dfrac{{ - 8}}{{15}}\\
 \Leftrightarrow x = \dfrac{1}{2}:\dfrac{{ - 8}}{{15}}\\
 \Leftrightarrow x = \dfrac{1}{2}.\dfrac{{ - 15}}{8}\\
 \Leftrightarrow x = \dfrac{{ - 15}}{{16}}\\
Vậy\,x = \dfrac{{ - 15}}{{16}}\\
c) - \dfrac{2}{3}:x + \dfrac{5}{8} = \dfrac{{ - 7}}{{12}}\\
 \Leftrightarrow  - \dfrac{2}{3}:x = \dfrac{{ - 7}}{{12}} - \dfrac{5}{8}\\
 \Leftrightarrow  - \dfrac{2}{3}:x = \dfrac{{ - 7.2 - 5.3}}{{24}}\\
 \Leftrightarrow  - \dfrac{2}{3}:x = \dfrac{{ - 29}}{{24}}\\
 \Leftrightarrow x = \dfrac{{ - 2}}{3}:\dfrac{{ - 29}}{{24}}\\
 \Leftrightarrow x = \dfrac{2}{3}.\dfrac{{24}}{{29}}\\
 \Leftrightarrow x = \dfrac{{16}}{{29}}\\
Vậy\,x = \dfrac{{16}}{{29}}\\
d)\dfrac{1}{2}x + 2\dfrac{1}{2} = 3\dfrac{1}{2}x - \dfrac{3}{4}\\
 \Leftrightarrow \dfrac{1}{2}x + \dfrac{5}{2} = \dfrac{7}{2}.x - \dfrac{3}{4}\\
 \Leftrightarrow 2x + 10 = 14x - 3\\
 \Leftrightarrow 12x = 13\\
 \Leftrightarrow x = \dfrac{{13}}{{12}}\\
Vậy\,x = \dfrac{{13}}{{12}}\\
e)\dfrac{2}{3}x - \dfrac{2}{5} = \dfrac{1}{2}x - \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{2}x = \dfrac{2}{5} - \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{1}{6}x = \dfrac{1}{{15}}\\
 \Leftrightarrow x = \dfrac{6}{{15}} = \dfrac{2}{5}\\
Vậy\,x = \dfrac{2}{5}\\
f)\dfrac{1}{3}x + \dfrac{2}{5}\left( {x + 1} \right) = 0\\
 \Leftrightarrow \dfrac{1}{3}x + \dfrac{2}{5}x + \dfrac{2}{5} = 0\\
 \Leftrightarrow \dfrac{{11}}{{15}}x = \dfrac{{ - 2}}{5}\\
 \Leftrightarrow x = \dfrac{{ - 2}}{5}.\dfrac{{15}}{{11}}\\
 \Leftrightarrow x = \dfrac{{ - 6}}{{11}}\\
Vậy\,x = \dfrac{{ - 6}}{{11}}\\
g)\dfrac{2}{3} - \dfrac{1}{3}\left( {x - \dfrac{3}{2}} \right) - \dfrac{1}{2}\left( {2x + 1} \right) = 5\\
 \Leftrightarrow \dfrac{2}{3} - \dfrac{1}{3}x + \dfrac{1}{2} - x - \dfrac{1}{2} = 5\\
 \Leftrightarrow x + \dfrac{1}{3}x = \dfrac{2}{3} - 5\\
 \Leftrightarrow \dfrac{4}{3}x = \dfrac{{ - 13}}{3}\\
 \Leftrightarrow x =  - \dfrac{{13}}{3}.\dfrac{3}{4}\\
 \Leftrightarrow x = \dfrac{{ - 13}}{4}\\
Vậy\,x = \dfrac{{ - 13}}{4}\\
h)\dfrac{{11}}{{15}} - \left( {\dfrac{7}{9} + x} \right).\dfrac{3}{8} = \dfrac{{61}}{{90}} + \dfrac{x}{3}\\
 \Leftrightarrow \dfrac{{11}}{{15}} - \dfrac{7}{{24}} - \dfrac{3}{8}x = \dfrac{{61}}{{90}} + \dfrac{x}{3}\\
 \Leftrightarrow \dfrac{x}{3} + \dfrac{3}{8}x = \dfrac{{11}}{{15}} - \dfrac{7}{{24}} - \dfrac{{61}}{{90}}\\
 \Leftrightarrow \dfrac{{17}}{{24}}x =  - \dfrac{{17}}{{72}}\\
 \Leftrightarrow x =  - \dfrac{1}{3}\\
Vậy\,x = \dfrac{{ - 1}}{3}
\end{array}$