Страница 107 номер 13.15, ГДЗ по алгебре за 10 класс к учебнику Мерзляка. Алгебра и начало математического анализа
Вычислите значение выражения: 1) \(\left(\frac{1}{16}\right)^{-\frac{3}{4}} + \left(\frac{1}{8}\right)^{-\frac{2}{3}} \cdot (0{,}81)^{-0{,}5}\); 2) \(16^{\frac{1}{8}} \cdot 8^{-\frac{5}{6}} \cdot 4^{1{,}5}\); 3) \(\frac{5^{\frac{3}{2}} \cdot 8^{\frac{1}{12}}}{9^{\frac{1}{6}}} \cdot \frac{8^{\frac{1}{4}}}{5^{\frac{5}{2}} \cdot 9^{\frac{1}{3}}}\); 4) \(\left(72^{\frac{2}{3}}\right)^{\frac{1}{2}} \cdot 2^{-\frac{4}{3}} : 36^{-\frac{1}{6}}\).
1) \[\begin{aligned} &\left(\frac{1}{16}\right)^{-\frac{3}{4}} + \left(\frac{1}{8}\right)^{-\frac{2}{3}} \cdot (0{,}81)^{-0{,}5} = {} \\ &= \left(2^{-4}\right)^{-\frac{3}{4}} + \left(2^{-3}\right)^{-\frac{2}{3}} \cdot \left(0{,}9^2\right)^{-0{,}5} = {} \\ &= 2^3 + 2^2 \cdot 0{,}9^{-1} = 8 + 4 \cdot \frac{10}{9} = 8 + \frac{40}{9} = 12\frac{4}{9}. \end{aligned}\]
2) \[\begin{aligned} &16^{\frac{1}{8}} \cdot 8^{-\frac{5}{6}} \cdot 4^{1{,}5} = \left(2^4\right)^{\frac{1}{8}} \cdot \left(2^3\right)^{-\frac{5}{6}} \cdot \left(2^2\right)^{1{,}5} = {} \\ &= 2^{\frac{1}{2}} \cdot 2^{-\frac{5}{2}} \cdot 2^3 = 2^{\frac{1}{2} - \frac{5}{2} + 3} = 2^1 = 2. \end{aligned}\]
3) \[\begin{aligned} &\frac{5^{\frac{3}{2}} \cdot 8^{\frac{1}{12}}}{9^{\frac{1}{6}}} \cdot \frac{8^{\frac{1}{4}}}{5^{\frac{5}{2}} \cdot 9^{\frac{1}{3}}} = 5^{\frac{3}{2} - \frac{5}{2}} \cdot 8^{\frac{1}{12} + \frac{1}{4}} \cdot 9^{-\frac{1}{6} - \frac{1}{3}} = {} \\ &= 5^{-1} \cdot 8^{\frac{1}{3}} \cdot 9^{-\frac{1}{2}} = \frac{2}{5 \cdot 3} = \frac{2}{15}. \end{aligned}\]
4) \[\begin{aligned} &\left(72^{\frac{2}{3}}\right)^{\frac{1}{2}} \cdot 2^{-\frac{4}{3}} : 36^{-\frac{1}{6}} = 72^{\frac{1}{3}} \cdot 2^{-\frac{4}{3}} \cdot 36^{\frac{1}{6}} = {} \\ &= \left(2^3 \cdot 3^2\right)^{\frac{1}{3}} \cdot 2^{-\frac{4}{3}} \cdot \left(2^2 \cdot 3^2\right)^{\frac{1}{6}} = 2 \cdot 3^{\frac{2}{3}} \cdot 2^{-\frac{4}{3}} \cdot 2^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} = {} \\ &= 2^{1 - \frac{4}{3} + \frac{1}{3}} \cdot 3^{\frac{2}{3} + \frac{1}{3}} = 2^0 \cdot 3 = 3. \end{aligned}\]
Ответ: 1) \(12\frac{4}{9}\); 2) \(2\); 3) \(\frac{2}{15}\); 4) \(3\).
Вычислите значение выражения: 1) \(\left(\frac{1}{16}\right)^{-\frac{3}{4}} + \left(\frac{1}{8}\right)^{-\frac{2}{3}} \cdot (0{,}81)^{-0{,}5}\); 2) \(16^{\frac{1}{8}} \cdot 8^{-\frac{5}{6}} \cdot 4^{1{,}5}\); 3) \(\frac{5^{\frac{3}{2}} \cdot 8^{\frac{1}{12}}}{9^{\frac{1}{6}}} \cdot \frac{8^{\frac{1}{4}}}{5^{\frac{5}{2}} \cdot 9^{\frac{1}{3}}}\); 4) \(\left(72^{\frac{2}{3}}\right)^{\frac{1}{2}} \cdot 2^{-\frac{4}{3}} : 36^{-\frac{1}{6}}\).
1) \[\begin{aligned} &\left(\frac{1}{16}\right)^{-\frac{3}{4}} + \left(\frac{1}{8}\right)^{-\frac{2}{3}} \cdot (0{,}81)^{-0{,}5} = {} \\ &= \left(2^{-4}\right)^{-\frac{3}{4}} + \left(2^{-3}\right)^{-\frac{2}{3}} \cdot \left(0{,}9^2\right)^{-0{,}5} = {} \\ &= 2^3 + 2^2 \cdot 0{,}9^{-1} = 8 + 4 \cdot \frac{10}{9} = 8 + \frac{40}{9} = 12\frac{4}{9}. \end{aligned}\]
2) \[\begin{aligned} &16^{\frac{1}{8}} \cdot 8^{-\frac{5}{6}} \cdot 4^{1{,}5} = \left(2^4\right)^{\frac{1}{8}} \cdot \left(2^3\right)^{-\frac{5}{6}} \cdot \left(2^2\right)^{1{,}5} = {} \\ &= 2^{\frac{1}{2}} \cdot 2^{-\frac{5}{2}} \cdot 2^3 = 2^{\frac{1}{2} - \frac{5}{2} + 3} = 2^1 = 2. \end{aligned}\]
3) \[\begin{aligned} &\frac{5^{\frac{3}{2}} \cdot 8^{\frac{1}{12}}}{9^{\frac{1}{6}}} \cdot \frac{8^{\frac{1}{4}}}{5^{\frac{5}{2}} \cdot 9^{\frac{1}{3}}} = 5^{\frac{3}{2} - \frac{5}{2}} \cdot 8^{\frac{1}{12} + \frac{1}{4}} \cdot 9^{-\frac{1}{6} - \frac{1}{3}} = {} \\ &= 5^{-1} \cdot 8^{\frac{1}{3}} \cdot 9^{-\frac{1}{2}} = \frac{2}{5 \cdot 3} = \frac{2}{15}. \end{aligned}\]
4) Деление на \(36^{-\frac{1}{6}}\) заменяем умножением на \(36^{\frac{1}{6}}\), так как \(\frac{1}{36^{-\frac{1}{6}}} = 36^{\frac{1}{6}}\). Имеем:
\[\begin{aligned} &\left(72^{\frac{2}{3}}\right)^{\frac{1}{2}} \cdot 2^{-\frac{4}{3}} : 36^{-\frac{1}{6}} = 72^{\frac{1}{3}} \cdot 2^{-\frac{4}{3}} \cdot 36^{\frac{1}{6}} = {} \\ &= \left(2^3 \cdot 3^2\right)^{\frac{1}{3}} \cdot 2^{-\frac{4}{3}} \cdot \left(2^2 \cdot 3^2\right)^{\frac{1}{6}} = 2 \cdot 3^{\frac{2}{3}} \cdot 2^{-\frac{4}{3}} \cdot 2^{\frac{1}{3}} \cdot 3^{\frac{1}{3}} = {} \\ &= 2^{1 - \frac{4}{3} + \frac{1}{3}} \cdot 3^{\frac{2}{3} + \frac{1}{3}} = 2^0 \cdot 3 = 3. \end{aligned}\]
Ответ: 1) \(12\frac{4}{9}\); 2) \(2\); 3) \(\frac{2}{15}\); 4) \(3\).
