Đáp án:

$\begin{array}{l}
B1)1)\\
\dfrac{{ - 2}}{{15}} + \dfrac{3}{{10}} = \dfrac{{ - 8}}{{60}} + \dfrac{{18}}{{60}} = \dfrac{{10}}{{60}} = \dfrac{1}{6}\\
2)\dfrac{{ - 8}}{7} - \dfrac{5}{{14}} = \dfrac{{ - 16}}{{14}} - \dfrac{5}{{14}} = \dfrac{{ - 21}}{{14}} = \dfrac{{ - 3}}{2}\\
3)\dfrac{{ - 4}}{{11}}.\dfrac{{55}}{{20}} = \dfrac{{ - 1}}{1}.\dfrac{5}{5} =  - 1\\
4)\dfrac{{ - 5}}{{20}}:\dfrac{{30}}{{ - 14}} + \dfrac{1}{{ - 3}}\\
 = \dfrac{{ - 5}}{{20}}.\dfrac{{ - 14}}{{30}} + \dfrac{{ - 1}}{3}\\
 = \dfrac{1}{{10}}.\dfrac{7}{6} + \dfrac{{ - 1}}{3}\\
 = \dfrac{7}{{60}} - \dfrac{{20}}{{60}}\\
 = \dfrac{{ - 13}}{{60}}\\
5)1\dfrac{1}{2}.\dfrac{{ - 10}}{9} = \dfrac{3}{2}.\dfrac{{ - 10}}{9} = \dfrac{{ - 5}}{3}\\
6)\left( {\dfrac{2}{5} - \dfrac{1}{2}} \right):\dfrac{{ - 7}}{{30}}\\
 = \dfrac{{ - 1}}{{10}}.\dfrac{{ - 30}}{7}\\
 = \dfrac{3}{7}\\
B2)a)\\
\dfrac{{12}}{{15}} + x = \dfrac{{ - 3}}{5}\\
 \Leftrightarrow x = \dfrac{{ - 7}}{5}\\
Vậy\,x = \dfrac{{ - 7}}{5}\\
b)\dfrac{4}{7}.x - 2 = \dfrac{{ - 1}}{3}\\
 \Leftrightarrow \dfrac{4}{7}x = \dfrac{5}{6}\\
 \Leftrightarrow x = \dfrac{{35}}{{24}}\\
Vậy\,x = \dfrac{{35}}{{24}}\\
c)\dfrac{8}{9} - \left( {\dfrac{2}{3} + x} \right) = \dfrac{5}{6}\\
 \Leftrightarrow \dfrac{8}{9} - \dfrac{2}{3} - x = \dfrac{5}{6}\\
 \Leftrightarrow x = \dfrac{{ - 11}}{{18}}\\
Vậy\,x = \dfrac{{ - 11}}{{18}}\\
d)\dfrac{4}{7} + \dfrac{3}{7}:x = \dfrac{2}{5}\\
 \Leftrightarrow \dfrac{3}{7}.\dfrac{1}{x} = \dfrac{{ - 6}}{{35}}\\
 \Leftrightarrow \dfrac{1}{x} = \dfrac{{ - 2}}{5}\\
 \Leftrightarrow x =  - \dfrac{5}{2}\\
Vậy\,x = \dfrac{{ - 5}}{2}\\
e)\dfrac{3}{{35}} + \left( {\dfrac{3}{5} - x} \right) = \dfrac{2}{7}\\
 \Leftrightarrow \dfrac{3}{{35}} + \dfrac{3}{5} - x = \dfrac{2}{7}\\
 \Leftrightarrow x = \dfrac{2}{5}\\
Vậy\,x = \dfrac{2}{5}\\
g)\left( {5x - 1} \right)\left( {2x + \dfrac{1}{3}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5x - 1 = 0\\
2x + \dfrac{1}{3} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x =  - \dfrac{1}{6}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{5},x = \dfrac{{ - 1}}{6}
\end{array}$