Страница 35 номер 95, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Вычислить: 1) \(\sqrt[3]{5^3 \cdot 7^3}\), \(\sqrt[4]{324} \cdot \sqrt[4]{4}\), \(\sqrt[4]{15\frac{5}{8}} : \sqrt[4]{\frac{2}{5}}\); 2) \(56^0 : 8^{-2}\), \(16^{\frac{1}{4}} \cdot 25^{\frac{1}{2}}\), \(\left(\frac{1}{15}\right)^{-1} : 9^{\frac{1}{2}}\), \(8^{\frac{1}{3}}\left(\frac{1}{2}\right)^4 : 16^{-1}\); 3) \(\frac{5^{\frac{1}{4}} \cdot 5^{-\frac{1}{4}}}{5^2}\), \(\frac{7^{\frac{7}{3}} \cdot 7^{-\frac{4}{3}}}{7^2}\), \(\frac{0{,}3^{0{,}3} \cdot 0{,}3^{-1}}{0{,}3^{1{,}3}}\).
1) \(\sqrt[3]{5^{3} \cdot 7^{3}} = 5 \cdot 7 = 35\);
\(\sqrt[4]{324} \cdot \sqrt[4]{4} = \sqrt[4]{3^{4} \cdot 2^{2} \cdot 2^{2}} = 3 \cdot 2 = 6;\)
\(\sqrt[4]{15\dfrac{5}{8}} : \sqrt[4]{\dfrac{2}{5}} = \sqrt[4]{\dfrac{125}{8} \cdot \dfrac{5}{2}} = \sqrt[4]{\dfrac{5^{4}}{2^{4}}} = \dfrac{5}{2} = 2{,}5;\)
2) \(56^{0} : 8^{-2} = 1 : \frac{1}{64} = 64\);
\(16^{\frac{1}{4}} \cdot 25^{\frac{1}{2}} = \sqrt[4]{2^{4}} \cdot \sqrt{25} = 2 \cdot 5 = 10;\)
\(\left(\dfrac{1}{15}\right)^{-1} : 9^{\frac{1}{2}} = 15 : \sqrt{9} = 15 : 3 = 5;\)
\(8^{\frac{1}{3}} \cdot \left(\dfrac{1}{2}\right)^{4} : 16^{-1} = \sqrt[3]{2^{3}} \cdot \dfrac{1}{16} : \dfrac{1}{16} = \sqrt[3]{2^{3}} = 2;\)
3) \(\frac{5^{\frac{1}{4}} \cdot 5^{-\frac{1}{4}}}{5^{2}} = 5^{\frac{1}{4} - \frac{1}{4} - 2} = 5^{-2} = \frac{1}{25}\);
\(\dfrac{7^{\frac{7}{3}} \cdot 7^{-\frac{4}{3}}}{7^{2}} = 7^{\frac{7}{3} - \frac{4}{3} - 2} = 7^{-1} = \dfrac{1}{7};\)
\(\begin{aligned} &\dfrac{0{,}3^{0{,}3} \cdot 0{,}3^{-1}}{0{,}3^{1{,}3}} = 0{,}3^{0{,}3 - 1 - 1{,}3} = 0{,}3^{-2} = \dfrac{1}{0{,}09} = {} \\ &= \dfrac{100}{9} = 11\dfrac{1}{9}. \end{aligned}\)
