Страница 31 номер 58, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить (57–60). 1) \(2^{\frac{4}{5}} \cdot 2^{\frac{11}{5}}\); 2) \(5^{\frac{2}{7}} \cdot 5^{\frac{5}{7}}\); 3) \(9^{\frac{2}{3}} : 9^{\frac{1}{6}}\); 4) \(4^{\frac{1}{3}} : 4^{\frac{5}{6}}\); 5) \(\left(8^{\frac{1}{12}}\right)^{-4}\).
1) \(2^{\frac{4}{5}} \cdot 2^{\frac{11}{5}} = \sqrt[5]{2^{4} \cdot 2^{11}} = \sqrt[5]{(2^{3})^{5}} = 2^{3} = 8\);
2) \(5^{\frac{2}{7}} \cdot 5^{\frac{5}{7}} = \sqrt[7]{5^{2} \cdot 5^{5}} = \sqrt[7]{5^{7}} = 5\);
3) \(9^{\frac{2}{3}} : 9^{\frac{1}{6}} = \sqrt[3]{9^{2}} : \sqrt[6]{9} = \sqrt[6]{9^{4} : 9} = \sqrt[6]{9^{3}} = 9^{\frac{3}{6}} = 9^{\frac{1}{2}} = \sqrt{9} = 3\);
4) \(4^{\frac{1}{3}} : 4^{\frac{5}{6}} = \sqrt[3]{4} : \sqrt[6]{4^{5}} = \sqrt[6]{4^{2} : 4^{5}} = \sqrt[6]{4^{-3}} = 4^{-\frac{3}{6}} = \frac{1}{4^{\frac{1}{2}}} = \frac{1}{\sqrt{4}} = \frac{1}{2}\);
5) \(\left(8^{\frac{1}{12}}\right)^{-4} = \frac{1}{\sqrt[12]{8^{4}}} = \frac{1}{\sqrt[12]{(2^{3})^{4}}} = \frac{1}{2}\).
