Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Задачи на повторение 29

\[\boxed{\mathbf{П.\ 29}}\]

\[\textbf{а)}\ a = 3:\]

\[3a^{3} = 3 \cdot 27 = 81.\]

\[a = - 3:\]

\[3a^{3} = 3 \cdot ( - 27) = - 81.\]

\[a = \frac{1}{3}:\]

\[3a^{3} = 3 \cdot \left( \frac{1}{3} \right)^{3} = 3 \cdot \frac{1}{27} = \frac{1}{9}.\]

\[\textbf{б)}\ a = 3:\]

\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot 9 = - 1.\]

\[a = - 3:\]

\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot 9 = - 1.\]

\[a = \frac{1}{3}:\]

\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot \left( \frac{1}{9} \right) = - \frac{1}{81}.\]

\[\textbf{в)}\ a = 3:\]

\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot 9 = 2 \cdot 3 = 6.\]

\[a = - 3:\]

\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot 9 = 2 \cdot 3 = 6.\]

\[a = \frac{1}{3}:\]

\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot \frac{1}{9} = \frac{2}{27}.\]

\[\textbf{г)}\ a = 3:\]

\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot 27 = - 9.\]

\[a = - 3:\]

\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot ( - 27) = 9.\]

\[a = \frac{1}{3}:\]

\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot \frac{1}{27} = - \frac{1}{81}.\]