Đáp án:
$\begin{array}{l}
8){\left( {\dfrac{1}{3}} \right)^{50}}.{\left( { - 9} \right)^{25}} - \dfrac{2}{3}:4\\
= \dfrac{1}{{{3^{50}}}}.{\left( { - {3^2}} \right)^{25}} - \dfrac{2}{3}.\dfrac{1}{4}\\
= - \dfrac{1}{{{3^{50}}}}{.3^{50}} - \dfrac{1}{6}\\
= - 1 - \dfrac{1}{6}\\
= - \dfrac{7}{6}\\
9)\dfrac{3}{5}:\left( { - \dfrac{1}{{15}} - \dfrac{1}{6}} \right) + \dfrac{3}{5}:\left( {\dfrac{{ - 1}}{3} - 1\dfrac{1}{{15}}} \right)\\
= \dfrac{3}{5}:\left( { - \dfrac{1}{{15}} - \dfrac{1}{6} - \dfrac{1}{3} - \dfrac{{16}}{{15}}} \right)\\
= \dfrac{3}{5}:\left( {\dfrac{{ - 17}}{{15}} - \dfrac{1}{2}} \right)\\
= \dfrac{3}{5}:\left( {\dfrac{{ - 49}}{{30}}} \right)\\
= \dfrac{3}{5}.\dfrac{{ - 30}}{{49}}\\
= - \dfrac{{18}}{{49}}\\
11)10\sqrt {0,01} .\sqrt {\dfrac{{16}}{9}} + 3\sqrt {49} - \dfrac{1}{6}\sqrt 4 \\
= 10.0,1.\dfrac{4}{3} + 3.7 - \dfrac{1}{6}.2\\
= \dfrac{4}{3} + 21 - \dfrac{1}{3}\\
= 1 + 21\\
= 22\\
12)\dfrac{{{2^4}{{.2}^6}}}{{{{\left( {{2^5}} \right)}^2}}} - \dfrac{{{2^5}{{.15}^3}}}{{{6^3}{{.10}^2}}}\\
= \dfrac{{{2^{10}}}}{{{2^{10}}}} - \dfrac{{{2^5}{{.3}^3}{{.5}^3}}}{{{2^3}{{.3}^3}{{.2}^2}{{.5}^2}}}\\
= 1 - 5\\
= - 4
\end{array}$
