Đáp án:
$\begin{array}{l}
c){\left( { - \dfrac{1}{3}} \right)^2}:\dfrac{{15}}{8} - \sqrt {\dfrac{{64}}{9}} .\dfrac{2}{{15}} + \left| {\dfrac{{10}}{{27}} - \dfrac{2}{3}} \right|\\
= \dfrac{1}{9}.\dfrac{8}{{15}} - \dfrac{8}{3}.\dfrac{2}{{15}} + \left| {\dfrac{{10 - 2.9}}{{27}}} \right|\\
= \dfrac{8}{{15}}.\left( {\dfrac{1}{9} - \dfrac{2}{3}} \right) + \dfrac{8}{{27}}\\
= \dfrac{8}{{15}}.\dfrac{{ - 5}}{9} + \dfrac{8}{{27}}\\
= \dfrac{{ - 8}}{{27}} + \dfrac{8}{{27}}\\
= 0\\
B2)\\
b)\dfrac{x}{{27}} = \dfrac{{ - 2}}{{3,6}}\\
\Leftrightarrow x = \dfrac{{ - 2}}{{3,6}}.27\\
\Leftrightarrow x = - \dfrac{{2.30}}{4} = - 15\\
Vậy\,x = - 15\\
c){1^{2020}} - {\left( {\dfrac{2}{3}x - 5} \right)^3} = \dfrac{{35}}{{27}}\\
\Leftrightarrow 1 - {\left( {\dfrac{2}{3}x - 5} \right)^3} = \dfrac{{35}}{{27}}\\
\Leftrightarrow {\left( {\dfrac{2}{3}x - 5} \right)^3} = 1 - \dfrac{{35}}{{27}}\\
\Leftrightarrow {\left( {\dfrac{2}{3}x - 5} \right)^3} = \dfrac{{ - 8}}{{27}}\\
\Leftrightarrow \left( {\dfrac{2}{3}x - 5} \right) = - \dfrac{2}{3}\\
\Leftrightarrow \dfrac{2}{3}x = - \dfrac{2}{3} + 5\\
\Leftrightarrow \dfrac{2}{3}x = \dfrac{{13}}{3}\\
\Leftrightarrow x = \dfrac{{13}}{3}.\dfrac{3}{2}\\
\Leftrightarrow x = \dfrac{{13}}{2}\\
Vậy\,x = \dfrac{{13}}{2}
\end{array}$
