Đáp án:
$\begin{array}{l}
5)a)A = \dfrac{5}{{12}} + \dfrac{4}{5}:\dfrac{{ - 3}}{4} - \dfrac{1}{4}\\
= \dfrac{5}{{12}} + \dfrac{4}{5}.\dfrac{{ - 4}}{3} - \dfrac{1}{4}\\
= \dfrac{5}{{12}} - \dfrac{{16}}{{15}} - \dfrac{1}{4}\\
= \dfrac{{5.5 - 16.4 - 15}}{{60}}\\
= \dfrac{{ - 54}}{{60}} = \dfrac{{ - 9}}{{10}}\\
b)B = {\left( {1 - \dfrac{3}{7}} \right)^2} + \left| {\dfrac{{ - 5}}{7}} \right| + \dfrac{{ - 4}}{{15}}\\
= {\left( {\dfrac{4}{7}} \right)^2} + \dfrac{5}{7} - \dfrac{4}{{15}}\\
= \dfrac{{21}}{7} - \dfrac{4}{{15}}\\
= 3 - \dfrac{4}{{15}}\\
= \dfrac{{41}}{{15}}\\
c)C = \dfrac{{{4^{20}}{{.3}^{35}}}}{{{2^{37}}{{.27}^{12}}}}\\
= \dfrac{{{2^{40}}{{.3}^{35}}}}{{{2^{37}}{{.3}^{36}}}} = \dfrac{{{2^3}}}{3} = \dfrac{8}{3}\\
d)D = {\left( {\dfrac{{ - 2}}{3}} \right)^2} - \left| {\dfrac{{ - 13}}{{15}}} \right| + \dfrac{5}{3}\\
= \dfrac{4}{9} - \dfrac{{13}}{{15}} + \dfrac{5}{3}\\
= \dfrac{{4.5 - 13.3 + 5.15}}{{45}}\\
= \dfrac{{56}}{{45}}\\
6)a)\dfrac{8}{9} - \dfrac{1}{9}x = \dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{9}x = \dfrac{8}{9} - \dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{9}x = \dfrac{{8 - 2.3}}{9} = \dfrac{2}{9}\\
\Leftrightarrow x = \dfrac{2}{9}:\dfrac{1}{9}\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
b)\left| {x + \dfrac{3}{4}} \right| - \dfrac{5}{6} = 3\dfrac{1}{5}\\
\Leftrightarrow \left| {x + \dfrac{3}{4}} \right| = \dfrac{{16}}{5} + \dfrac{5}{6}\\
\Leftrightarrow \left| {x + \dfrac{3}{4}} \right| = \dfrac{{121}}{{30}}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{4} = \dfrac{{121}}{{30}}\\
x + \dfrac{3}{4} = - \dfrac{{121}}{{30}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{121}}{{30}} - \dfrac{3}{4} = \dfrac{{197}}{{60}}\\
x = \dfrac{{ - 121}}{{30}} - \dfrac{3}{4} = \dfrac{{ - 287}}{{60}}
\end{array} \right.\\
Vậy\,x = \dfrac{{197}}{{60}};x = \dfrac{{ - 287}}{{60}}
\end{array}$
