Страница 35 номер 94, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить: 1) \(48^0\), \(10^{-2}\), \(\left(\frac{2}{3}\right)^{-1}\), \((0{,}3)^{-3}\), \((-1{,}2)^{-2}\), \(\left(2\frac{1}{4}\right)^{-2}\); 2) \(\sqrt[3]{27}\), \(\sqrt[4]{81}\), \(\sqrt[5]{32}\), \(\sqrt[6]{8^2}\), \(\sqrt[8]{16^2}\), \(\sqrt[3]{27^2}\); 3) \(8^{\frac{1}{3}}\), \(27^{\frac{2}{3}}\), \(10\,000^{\frac{1}{4}}\), \(32^{\frac{2}{5}}\), \(32^{-\frac{3}{5}}\), \(\left(\frac{27}{64}\right)^{\frac{2}{3}}\).
1) \(48^{0} = 1\);
\(10^{-2} = \dfrac{1}{10^{2}} = 0{,}01;\)
\(\left(\dfrac{2}{3}\right)^{-1} = \dfrac{3}{2} = 1{,}5;\)
\((0{,}3)^{-3} = \left(\dfrac{10}{3}\right)^{3} = \dfrac{1000}{27} = 37\dfrac{1}{27};\)
\((-1{,}2)^{-2} = \left(-\dfrac{10}{12}\right)^{2} = \dfrac{100}{144} = \dfrac{25}{36};\)
\(\left(2\dfrac{1}{4}\right)^{-2} = \left(\dfrac{9}{4}\right)^{-2} = \dfrac{16}{81};\)
2) \(\sqrt[3]{27} = \sqrt[3]{3^{3}} = 3\);
\(\sqrt[4]{81} = \sqrt[4]{3^{4}} = 3;\)
\(\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2;\)
\(\sqrt[6]{8^{2}} = \sqrt[6]{(2^{3})^{2}} = \sqrt[6]{2^{6}} = 2;\)
\(\sqrt[8]{16^{2}} = \sqrt[8]{(2^{4})^{2}} = \sqrt[8]{2^{8}} = 2;\)
\(\sqrt[3]{27^{2}} = \sqrt[3]{(3^{3})^{2}} = 3^{2} = 9;\)
3) \(8^{\frac{1}{3}} = \sqrt[3]{2^{3}} = 2\);
\(27^{\frac{2}{3}} = \sqrt[3]{(3^{3})^{2}} = 9;\)
\(10000^{\frac{1}{4}} = \sqrt[4]{10^{4}} = 10;\)
\(32^{\frac{2}{5}} = \sqrt[5]{(2^{5})^{2}} = 2^{2} = 4;\)
\(32^{-\frac{3}{5}} = \dfrac{1}{\sqrt[5]{(2^{5})^{3}}} = \dfrac{1}{2^{3}} = \dfrac{1}{8};\)
\(\left(\dfrac{27}{64}\right)^{\frac{2}{3}} = \sqrt[3]{\left(\dfrac{3^{3}}{4^{3}}\right)^{2}} = \left(\dfrac{3}{4}\right)^{2} = \dfrac{9}{16}.\)
