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МатематикаРешебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. Задачи на повторение 29
29
\[\boxed{\mathbf{П.\ 29}}\]
\[\textbf{а)}\ a = 3:\]
\[3a^{3} = 3 \cdot 27 = 81.\]
\[a = - 3:\]
\[3a^{3} = 3 \cdot ( - 27) = - 81.\]
\[a = \frac{1}{3}:\]
\[3a^{3} = 3 \cdot \left( \frac{1}{3} \right)^{3} = 3 \cdot \frac{1}{27} = \frac{1}{9}.\]
\[\textbf{б)}\ a = 3:\]
\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot 9 = - 1.\]
\[a = - 3:\]
\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot 9 = - 1.\]
\[a = \frac{1}{3}:\]
\[- \frac{1}{9}a^{2} = - \frac{1}{9} \cdot \left( \frac{1}{9} \right) = - \frac{1}{81}.\]
\[\textbf{в)}\ a = 3:\]
\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot 9 = 2 \cdot 3 = 6.\]
\[a = - 3:\]
\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot 9 = 2 \cdot 3 = 6.\]
\[a = \frac{1}{3}:\]
\[\frac{2}{3}a^{2} = \frac{2}{3} \cdot \frac{1}{9} = \frac{2}{27}.\]
\[\textbf{г)}\ a = 3:\]
\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot 27 = - 9.\]
\[a = - 3:\]
\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot ( - 27) = 9.\]
\[a = \frac{1}{3}:\]
\[- \frac{1}{3}a^{3} = - \frac{1}{3} \cdot \frac{1}{27} = - \frac{1}{81}.\]
