ГДЗ по алгебре 9 класс Макарычев Задание 192

\[\boxed{\text{192\ (192).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 27^{\frac{1}{3}} = \sqrt[3]{27} = 3;\]

\[\textbf{б)}\ 25^{- \frac{1}{2}} = \frac{1}{\sqrt{25}} = \frac{1}{5};\ \]

\[\textbf{в)}\ {0,16}^{\frac{3}{2}} = \left( \sqrt{0,16} \right)^{3} =\]

\[= {0,4}^{3} = 0,064;\]

\[\textbf{г)}\ {0,64}^{- 1,5} = (0,64)^{- \frac{3}{2}} =\]

\[= \frac{1}{\left( \sqrt{0,64} \right)^{3}} = \frac{1}{{0,8}^{3}} = \frac{1}{0,512} =\]

\[= \frac{1000}{512} = 1\frac{61}{64};\]

\[\textbf{д)}\ 5 \cdot 32^{\frac{1}{5}} = 5 \cdot \sqrt[5]{32} = 5 \cdot 2 = 10;\]

\[\textbf{е)} - 64^{\frac{1}{3}} = - \sqrt[3]{64} = - 4;\]

\[\textbf{ж)}\ 6 \cdot 8^{- \frac{1}{3}} = \frac{6}{\sqrt[3]{8}} = \frac{6}{2} = 3;\]

\[\textbf{з)}\ 7 \cdot {0,04}^{- \frac{1}{2}} = 7 \cdot \frac{1}{\sqrt{0,04}} =\]

\[= 7 \cdot \frac{1}{0,2} = 7 \cdot 5 = 35.\]