Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\dfrac{1}{2} - \left( {x + \dfrac{1}{3}} \right) = \dfrac{5}{6}\\
 \Leftrightarrow x + \dfrac{1}{3} = \dfrac{1}{2} - \dfrac{5}{6}\\
 \Leftrightarrow x + \dfrac{1}{3} =  - \dfrac{1}{3}\\
 \Leftrightarrow x =  - \dfrac{1}{3} - \dfrac{1}{3}\\
 \Leftrightarrow x =  - \dfrac{2}{3}\\
2,\\
\left( {\dfrac{3}{8} - \dfrac{1}{5}} \right) + \left( {\dfrac{5}{8} - x} \right) = \dfrac{1}{5}\\
 \Leftrightarrow \dfrac{7}{{40}} + \left( {\dfrac{5}{8} - x} \right) = \dfrac{1}{5}\\
 \Leftrightarrow \dfrac{5}{8} - x = \dfrac{1}{5} - \dfrac{7}{{40}}\\
 \Leftrightarrow \dfrac{5}{8} - x = \dfrac{1}{{40}}\\
 \Leftrightarrow x = \dfrac{5}{8} - \dfrac{1}{{40}}\\
 \Leftrightarrow x = \dfrac{3}{5}\\
3,\\
\dfrac{3}{2} - \left( {2x + \dfrac{4}{5}} \right) = \dfrac{2}{7}\\
 \Leftrightarrow 2x + \dfrac{4}{5} = \dfrac{3}{2} - \dfrac{2}{7}\\
 \Leftrightarrow 2x + \dfrac{4}{5} = \dfrac{{17}}{{14}}\\
 \Leftrightarrow 2x = \dfrac{{17}}{{14}} - \dfrac{4}{5}\\
 \Leftrightarrow 2x = \dfrac{{29}}{{70}}\\
 \Leftrightarrow x = \dfrac{{29}}{{140}}\\
4,\\
\left( {\dfrac{3}{5} - \dfrac{4}{3}} \right) + \left( {\dfrac{5}{8} - x} \right) = \dfrac{9}{7}\\
 \Leftrightarrow  - \dfrac{{11}}{{15}} + \left( {\dfrac{5}{8} - x} \right) = \dfrac{9}{7}\\
 \Leftrightarrow \dfrac{5}{8} - x = \dfrac{9}{7} + \dfrac{{11}}{{15}}\\
 \Leftrightarrow \dfrac{5}{8} - x = \dfrac{{212}}{{105}}\\
 \Leftrightarrow x = \dfrac{5}{8} - \dfrac{{212}}{{105}}\\
 \Leftrightarrow x =  - \dfrac{{1171}}{{840}}\\
5,\\
\dfrac{1}{3} + \left( {3x + \dfrac{2}{5}} \right) = \dfrac{5}{7}\\
 \Leftrightarrow 3x + \dfrac{2}{5} = \dfrac{5}{7} - \dfrac{1}{3}\\
 \Leftrightarrow 3x + \dfrac{2}{5} = \dfrac{8}{{21}}\\
 \Leftrightarrow 3x = \dfrac{8}{{21}} - \dfrac{2}{5}\\
 \Leftrightarrow 3x =  - \dfrac{2}{{105}}\\
 \Leftrightarrow x =  - \dfrac{2}{{315}}\\
6,\\
\left( {\dfrac{2}{5} - \dfrac{2}{3}} \right) + \left( {\dfrac{3}{7} - 2x} \right) = \dfrac{4}{9}\\
 \Leftrightarrow  - \dfrac{4}{{15}} + \left( {\dfrac{3}{7} - 2x} \right) = \dfrac{4}{9}\\
 \Leftrightarrow \dfrac{3}{7} - 2x = \dfrac{4}{9} + \dfrac{4}{{15}}\\
 \Leftrightarrow \dfrac{3}{7} - 2x = \dfrac{{32}}{{45}}\\
 \Leftrightarrow 2x = \dfrac{3}{7} - \dfrac{{32}}{{45}}\\
 \Leftrightarrow 2x =  - \dfrac{{89}}{{315}}\\
 \Leftrightarrow x =  - \dfrac{{89}}{{630}}
\end{array}\)

\(\begin{array}{l}
7,\\
\dfrac{2}{3} - \left( {4x + \dfrac{5}{6}} \right) = \dfrac{6}{5}\\
 \Leftrightarrow 4x + \dfrac{5}{6} = \dfrac{2}{3} - \dfrac{6}{5}\\
 \Leftrightarrow 4x + \dfrac{5}{6} =  - \dfrac{8}{{15}}\\
 \Leftrightarrow 4x =  - \dfrac{8}{{15}} - \dfrac{5}{6}\\
 \Leftrightarrow 4x =  - \dfrac{{41}}{{30}}\\
 \Leftrightarrow x =  - \dfrac{{41}}{{120}}\\
8,\\
\left( {\dfrac{5}{6} + \dfrac{4}{7}} \right) - \left( {\dfrac{3}{2} + 3x} \right) =  - \dfrac{1}{6}\\
 \Leftrightarrow \dfrac{{59}}{{42}} - \left( {\dfrac{3}{2} + 3x} \right) =  - \dfrac{1}{6}\\
 \Leftrightarrow \dfrac{3}{2} + 3x = \dfrac{{59}}{{42}} + \dfrac{1}{6}\\
 \Leftrightarrow \dfrac{3}{2} + 3x = \dfrac{{11}}{7}\\
 \Leftrightarrow 3x = \dfrac{{11}}{7} - \dfrac{3}{2}\\
 \Leftrightarrow 3x = \dfrac{1}{{14}}\\
 \Leftrightarrow x = \dfrac{1}{{42}}
\end{array}\)