Giải thích các bước giải:

 $\begin{array}{l} 1,\\ a,\\ {\left( {\dfrac{1}{2}} \right)^2} + \left| { - \dfrac{3}{4}} \right| - \sqrt {\dfrac{{81}}{{25}}} .\dfrac{{10}}{3}\\  = \dfrac{1}{4} + \dfrac{3}{4} - \sqrt {{{\left( {\dfrac{9}{5}} \right)}^2}} .\dfrac{{10}}{3}\\  = \dfrac{{1 + 3}}{4} - \dfrac{9}{5}.\dfrac{{10}}{3}\\  = 1 - 6 =  - 5\\ b,\\ \dfrac{{11}}{2}.4\dfrac{6}{3} - 2\dfrac{5}{3}.\dfrac{{11}}{2}\\  = \dfrac{{11}}{2}.\left( {4\dfrac{6}{3} - 2\dfrac{5}{3}} \right)\\  = \dfrac{{11}}{2}.\left( {4 + 2 - \dfrac{{11}}{3}} \right)\\  = \dfrac{{11}}{2}.\left( {6 - \dfrac{{11}}{3}} \right)\\  = \dfrac{{11}}{2}.\dfrac{7}{3} = \dfrac{{77}}{6}\\ c,\\ 16\dfrac{2}{7}:\left( { - \dfrac{3}{5}} \right) - 28\dfrac{2}{7}:\left( { - \dfrac{3}{5}} \right)\\  = \left( {16\dfrac{2}{7} - 28\dfrac{2}{7}} \right):\left( { - \dfrac{3}{5}} \right)\\  = \left( { - 12} \right):\left( { - \dfrac{3}{5}} \right)\\  = 12:\dfrac{3}{5} = 12.\dfrac{5}{3} = 4.5 = 20\\ d,\\ \dfrac{2}{3} - \dfrac{5}{3}:\dfrac{4}{3} + 25\% \\  = \dfrac{2}{3} - \dfrac{5}{3}.\dfrac{3}{4} + \dfrac{{25}}{{100}}\\  = \dfrac{2}{3} - \dfrac{5}{4} + \dfrac{1}{4}\\  = \dfrac{2}{3} - \dfrac{{5 - 1}}{4} = \dfrac{2}{3} - 1 =  - \dfrac{1}{3}\\ 2,\\ a,\,\,\,\left| {x - \dfrac{1}{4}} \right| + \dfrac{1}{2} = \dfrac{5}{8}\\  \Leftrightarrow \left| {x - \dfrac{1}{4}} \right| = \dfrac{5}{8} - \dfrac{1}{2}\\  \Leftrightarrow \left| {x - \dfrac{1}{4}} \right| = \dfrac{1}{8}\\  \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{1}{4} = \dfrac{1}{8}\\ x - \dfrac{1}{4} =  - \dfrac{1}{8} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{3}{8}\\ x = \dfrac{1}{8} \end{array} \right.\\ b,\,\,\,\, - 1,5 + \dfrac{{15}}{2}x = \dfrac{3}{4}\\  \Leftrightarrow \dfrac{{15}}{2}x = \dfrac{3}{4} + 1,5\\  \Leftrightarrow \dfrac{{15}}{2}x = \dfrac{9}{4}\\  \Leftrightarrow x = \dfrac{9}{4}:\dfrac{{15}}{2}\\  \Leftrightarrow x = \dfrac{3}{{10}}\\ c,\,\,\,\,\,\dfrac{5}{{12}}x =  - \dfrac{{15}}{4}\\  \Leftrightarrow x = \left( { - \dfrac{{15}}{4}} \right):\dfrac{5}{{12}}\\  \Leftrightarrow x = \left( { - \dfrac{{15}}{4}} \right).\dfrac{{12}}{5}\\  \Leftrightarrow x =  - 9\\ d,\\ \dfrac{3}{2} - \left| x \right| =  - \dfrac{5}{6}\\  \Leftrightarrow \left| x \right| = \dfrac{3}{2} - \left( { - \dfrac{5}{6}} \right)\\  \Leftrightarrow \left| x \right| = \dfrac{7}{3} \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{7}{3}\\ x =  - \dfrac{7}{3} \end{array} \right. \end{array}$

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {\dfrac{1}{2}} \right)^2} + \left| { - \dfrac{3}{4}} \right| - \sqrt {\dfrac{{81}}{{25}}} .\dfrac{{10}}{3}\\
 = \dfrac{1}{4} + \dfrac{3}{4} - \sqrt {{{\left( {\dfrac{9}{5}} \right)}^2}} .\dfrac{{10}}{3}\\
 = \dfrac{{1 + 3}}{4} - \dfrac{9}{5}.\dfrac{{10}}{3}\\
 = 1 - 6 =  - 5\\
b,\\
\dfrac{{11}}{2}.4\dfrac{6}{3} - 2\dfrac{5}{3}.\dfrac{{11}}{2}\\
 = \dfrac{{11}}{2}.\left( {4\dfrac{6}{3} - 2\dfrac{5}{3}} \right)\\
 = \dfrac{{11}}{2}.\left( {4 + 2 - \dfrac{{11}}{3}} \right)\\
 = \dfrac{{11}}{2}.\left( {6 - \dfrac{{11}}{3}} \right)\\
 = \dfrac{{11}}{2}.\dfrac{7}{3} = \dfrac{{77}}{6}\\
c,\\
16\dfrac{2}{7}:\left( { - \dfrac{3}{5}} \right) - 28\dfrac{2}{7}:\left( { - \dfrac{3}{5}} \right)\\
 = \left( {16\dfrac{2}{7} - 28\dfrac{2}{7}} \right):\left( { - \dfrac{3}{5}} \right)\\
 = \left( { - 12} \right):\left( { - \dfrac{3}{5}} \right)\\
 = 12:\dfrac{3}{5} = 12.\dfrac{5}{3} = 4.5 = 20\\
d,\\
\dfrac{2}{3} - \dfrac{5}{3}:\dfrac{4}{3} + 25\% \\
 = \dfrac{2}{3} - \dfrac{5}{3}.\dfrac{3}{4} + \dfrac{{25}}{{100}}\\
 = \dfrac{2}{3} - \dfrac{5}{4} + \dfrac{1}{4}\\
 = \dfrac{2}{3} - \dfrac{{5 - 1}}{4} = \dfrac{2}{3} - 1 =  - \dfrac{1}{3}\\
2,\\
a,\,\,\,\left| {x - \dfrac{1}{4}} \right| + \dfrac{1}{2} = \dfrac{5}{8}\\
 \Leftrightarrow \left| {x - \dfrac{1}{4}} \right| = \dfrac{5}{8} - \dfrac{1}{2}\\
 \Leftrightarrow \left| {x - \dfrac{1}{4}} \right| = \dfrac{1}{8}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{4} = \dfrac{1}{8}\\
x - \dfrac{1}{4} =  - \dfrac{1}{8}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{8}\\
x = \dfrac{1}{8}
\end{array} \right.\\
b,\,\,\,\, - 1,5 + \dfrac{{15}}{2}x = \dfrac{3}{4}\\
 \Leftrightarrow \dfrac{{15}}{2}x = \dfrac{3}{4} + 1,5\\
 \Leftrightarrow \dfrac{{15}}{2}x = \dfrac{9}{4}\\
 \Leftrightarrow x = \dfrac{9}{4}:\dfrac{{15}}{2}\\
 \Leftrightarrow x = \dfrac{3}{{10}}\\
c,\,\,\,\,\,\dfrac{5}{{12}}x =  - \dfrac{{15}}{4}\\
 \Leftrightarrow x = \left( { - \dfrac{{15}}{4}} \right):\dfrac{5}{{12}}\\
 \Leftrightarrow x = \left( { - \dfrac{{15}}{4}} \right).\dfrac{{12}}{5}\\
 \Leftrightarrow x =  - 9\\
d,\\
\dfrac{3}{2} - \left| x \right| =  - \dfrac{5}{6}\\
 \Leftrightarrow \left| x \right| = \dfrac{3}{2} - \left( { - \dfrac{5}{6}} \right)\\
 \Leftrightarrow \left| x \right| = \dfrac{7}{3} \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x =  - \dfrac{7}{3}
\end{array} \right.
\end{array}\)