Страница 107 номер 13.16, ГДЗ по алгебре за 10 класс к учебнику Мерзляка. Алгебра и начало математического анализа

1) \[\begin{aligned} &\left(343^{\frac{1}{2}} \cdot \left(\frac{1}{49}\right)^{\frac{3}{8}}\right)^{\frac{4}{3}} = \left(\left(7^3\right)^{\frac{1}{2}} \cdot \left(7^{-2}\right)^{\frac{3}{8}}\right)^{\frac{4}{3}} = {} \\ &= \left(7^{\frac{3}{2}} \cdot 7^{-\frac{3}{4}}\right)^{\frac{4}{3}} = \left(7^{\frac{3}{4}}\right)^{\frac{4}{3}} = 7^1 = 7. \end{aligned}\]

2) \[\begin{aligned} &10^{\frac{1}{4}} \cdot 40^{\frac{1}{4}} \cdot 5^{\frac{1}{2}} = (10 \cdot 40)^{\frac{1}{4}} \cdot 5^{\frac{1}{2}} = \left(20^2\right)^{\frac{1}{4}} \cdot 5^{\frac{1}{2}} = {} \\ &= 20^{\frac{1}{2}} \cdot 5^{\frac{1}{2}} = (20 \cdot 5)^{\frac{1}{2}} = 100^{\frac{1}{2}} = 10. \end{aligned}\]

3) \[\begin{aligned} &\frac{32^{0{,}24} \cdot 4^{0{,}7}}{64^{0{,}6} \cdot 16^{0{,}25}} = \frac{\left(2^5\right)^{0{,}24} \cdot \left(2^2\right)^{0{,}7}}{\left(2^6\right)^{0{,}6} \cdot \left(2^4\right)^{0{,}25}} = {} \\ &= \frac{2^{1{,}2} \cdot 2^{1{,}4}}{2^{3{,}6} \cdot 2^1} = 2^{2{,}6 - 4{,}6} = 2^{-2} = \frac{1}{4}. \end{aligned}\]

4) \[\begin{aligned} &\frac{12^{\frac{1}{2}}}{7^{\frac{2}{3}} \cdot 8^{-\frac{1}{6}}} \cdot \frac{3^{\frac{1}{2}} \cdot 7^{\frac{5}{3}}}{8^{\frac{1}{2}}} = \frac{12^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} \cdot 7^{\frac{5}{3}} \cdot 8^{\frac{1}{6}}}{7^{\frac{2}{3}} \cdot 8^{\frac{1}{2}}} = {} \\ &= (12 \cdot 3)^{\frac{1}{2}} \cdot 7^{\frac{5}{3} - \frac{2}{3}} \cdot 8^{\frac{1}{6} - \frac{1}{2}} = 36^{\frac{1}{2}} \cdot 7 \cdot 8^{-\frac{1}{3}} = 6 \cdot 7 \cdot \frac{1}{2} = 21. \end{aligned}\]

Ответ: 1) \(7\); 2) \(10\); 3) \(\frac{1}{4}\); 4) \(21\).