Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 94

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\[{1)\ 48}^{0} = 1;\]

\[10^{- 2} = \frac{1}{10^{2}} = \frac{1}{100} = 0,01;\]

\[\left( \frac{2}{3} \right)^{- 1} = \frac{3}{2} = 1,5;\]

\[(0,3)^{- 3} = \left( \frac{3}{10} \right)^{- 3} = \left( \frac{10}{3} \right)^{3} =\]

\[= \frac{1000}{27} = 37\frac{1}{27};\]

\[( - 1,2)^{- 2} = \left( - \frac{12}{10} \right)^{- 2} =\]

\[= \left( - \frac{6}{5} \right)^{- 2} = \left( - \frac{5}{6} \right)^{2} = \frac{25}{36};\]

\[\left( 2\frac{1}{4} \right)^{- 2} = \left( \frac{2 \bullet 4 + 1}{4} \right)^{- 2} =\]

\[= \left( \frac{9}{4} \right)^{- 2} = \left( \frac{4}{9} \right)^{2} = \frac{16}{81}\text{.\ }\]

\[2)\ \sqrt[3]{27} = \sqrt[3]{3^{3}} = 3;\]

\[\sqrt[4]{81} = \sqrt[4]{3^{4}} = 3;\]

\[\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2;\]

\[\sqrt[8]{16^{2}} = \sqrt[8]{\left( 2^{4} \right)^{2}} = \sqrt[8]{2^{8}} = 2;\]

\[\sqrt[3]{27^{2}} = \sqrt[3]{\left( 3^{3} \right)^{2}} = \sqrt[3]{3^{3 \bullet 2}} =\]

\[= 3^{2} = 9.\]

\[3)\ 8^{\frac{1}{3}} = \left( 2^{3} \right)^{\frac{1}{3}} = 2;\]

\[27^{\frac{2}{3}} = \left( 3^{3} \right)^{\frac{2}{3}} = 3^{2} = 9;\]

\[10\ 000^{\frac{1}{4}} = \left( 10^{4} \right)^{\frac{1}{4}} = 10;\]

\[32^{\frac{2}{5}} = \left( 2^{5} \right)^{\frac{2}{5}} = 2^{2} = 4;\]

\[32^{- \frac{3}{5}} = \left( 2^{5} \right)^{- \frac{3}{5}} = 2^{- 3} = \frac{1}{2^{3}} = \frac{1}{8};\]

\[\left( \frac{27}{64} \right)^{\frac{2}{3}} = \left( \frac{3^{3}}{4^{3}} \right)^{\frac{2}{3}} = \frac{3^{2}}{4^{2}} = \frac{9}{16}.\]