\[\boxed{\text{116\ (116).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left( \frac{2a}{p^{2}q^{3}} \right)^{4} = \frac{(2a)^{4}}{\left( p^{2}q^{3} \right)^{4}} = \frac{16a^{4}}{p^{8}q^{12}}\ \]
\[\textbf{б)}\ \left( \frac{3a^{2}b^{3}}{s^{4}} \right)^{2} = \frac{\left( 3a^{2}b^{3} \right)^{2}}{\left( s^{4} \right)^{2}} =\]
\[= \frac{9a^{4}b^{6}}{s^{8}}\]
\[\textbf{в)}\ \left( - \frac{2a^{2}b}{3mn^{3}} \right)^{2} = \frac{\left( 2a^{2}b \right)^{2}}{\left( 3mn^{3} \right)^{2}} =\]
\[= \frac{4a^{4}b^{2}}{9m^{2}n^{6}}\]
\[\textbf{г)}\ \left( - \frac{3x^{2}}{2y^{3}} \right)^{3} = - \frac{\left( 3x^{2} \right)^{3}}{\left( 2y^{3} \right)^{3}} =\]
\[= - \frac{27x^{6}}{8y^{9}}\]