Помоги вычислить выражения (57-60): 1) 64^(2/3)

Сейчас помогу тебе разобраться с этими заданиями. Тут нужно вспомнить свойства степеней и корней, чтобы упростить выражения. **Задание 1** 1) $64^\frac{2}{3} = (4^3)^\frac{2}{3} = 4^{3 \cdot \frac{2}{3}} = 4^2 = 16$ 2) $27^\frac{1}{3} = (3^3)^\frac{1}{3} = 3^{3 \cdot \frac{1}{3}} = 3^1 = 3$ 3) $8^\frac{2}{3} = (2^3)^\frac{2}{3} = 2^{3 \cdot \frac{2}{3}} = 2^2 = 4$ 4) $81^\frac{1}{4} = (3^4)^\frac{1}{4} = 3^{4 \cdot \frac{1}{4}} = 3^1 = 3$ 5) $16^{-0.75} = 16^{-\frac{3}{4}} = (2^4)^{-\frac{3}{4}} = 2^{4 \cdot (-\frac{3}{4})} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$ 6) $9^{1.5} = 9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}} = 3^{2 \cdot \frac{3}{2}} = 3^3 = 27$ **Задание 2** 1) $2^\frac{6}{11} \cdot 2^\frac{5}{11} = 2^{\frac{6}{11} + \frac{5}{11}} = 2^{\frac{11}{11}} = 2^1 = 2$ 2) $5^\frac{2}{7} \cdot 5^\frac{5}{7} = 5^{\frac{2}{7} + \frac{5}{7}} = 5^{\frac{7}{7}} = 5^1 = 5$ 3) $9^\frac{2}{3} : 9^\frac{1}{6} = 9^{\frac{2}{3} - \frac{1}{6}} = 9^{\frac{4}{6} - \frac{1}{6}} = 9^{\frac{3}{6}} = 9^{\frac{1}{2}} = \sqrt{9} = 3$ 4) $4^\frac{1}{8} : 4^\frac{5}{8} = 4^{\frac{1}{8} - \frac{5}{8}} = 4^{-\frac{4}{8}} = 4^{-\frac{1}{2}} = \frac{1}{4^\frac{1}{2}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$ 5) $(8^\frac{1}{2})^{-4} = 8^{\frac{1}{2} \cdot (-4)} = 8^{-2} = \frac{1}{8^2} = \frac{1}{64}$ **Задание 3** 1) $9^\frac{3}{2} \cdot 27^\frac{2}{3} = (3^2)^\frac{3}{2} \cdot (3^3)^\frac{2}{3} = 3^{2 \cdot \frac{3}{2}} \cdot 3^{3 \cdot \frac{2}{3}} = 3^3 \cdot 3^2 = 3^{3+2} = 3^5 = 243$ 2) $7^\frac{2}{3} \cdot 49^\frac{3}{4} = 7^\frac{2}{3} \cdot (7^2)^\frac{3}{4} = 7^\frac{2}{3} \cdot 7^{2 \cdot \frac{3}{4}} = 7^\frac{2}{3} \cdot 7^\frac{3}{2} = 7^{\frac{2}{3} + \frac{3}{2}} = 7^{\frac{4}{6} + \frac{9}{6}} = 7^{\frac{13}{6}}$ 3) $144^\frac{3}{4} : 9^\frac{3}{2} = (12^2)^\frac{3}{4} : (3^2)^\frac{3}{2} = 12^{2 \cdot \frac{3}{4}} : 3^{2 \cdot \frac{3}{2}} = 12^\frac{3}{2} : 3^3 = (4 \cdot 3)^\frac{3}{2} : 27 = 4^\frac{3}{2} \cdot 3^\frac{3}{2} : 27 = (2^2)^\frac{3}{2} \cdot 3^\frac{3}{2} : 27 = 2^3 \cdot 3^\frac{3}{2} : 27 = 8 \cdot 3^\frac{3}{2} : 27 = 8 \cdot \sqrt{3^3} : 27 = 8 \cdot \sqrt{27} : 27 = \frac{8 \sqrt{27}}{27} = \frac{8 \sqrt{9 \cdot 3}}{27} = \frac{8 \cdot 3 \sqrt{3}}{27} = \frac{24 \sqrt{3}}{27} = \frac{8 \sqrt{3}}{9}$ 4) $150^\frac{2}{3} : 6^\frac{2}{3} = (150:6)^\frac{2}{3} = 25^\frac{2}{3} = (5^2)^\frac{2}{3} = 5^{\frac{4}{3}}$ **Задание 4** 1) $(\frac{1}{16})^{-0.75} + (\frac{1}{8})^{- \frac{4}{3}} = (\frac{1}{2^4})^{- \frac{3}{4}} + (\frac{1}{2^3})^{- \frac{4}{3}} = (2^{-4})^{- \frac{3}{4}} + (2^{-3})^{- \frac{4}{3}} = 2^{(-4) \cdot (- \frac{3}{4})} + 2^{(-3) \cdot (- \frac{4}{3})} = 2^3 + 2^4 = 8 + 16 = 24$ 2) $(0.04)^{-1.5} - (0.125)^{- \frac{2}{3}} = (\frac{4}{100})^{- \frac{3}{2}} - (\frac{125}{1000})^{- \frac{2}{3}} = (\frac{1}{25})^{- \frac{3}{2}} - (\frac{1}{8})^{- \frac{2}{3}} = (25)^\frac{3}{2} - (8)^\frac{2}{3} = (5^2)^\frac{3}{2} - (2^3)^\frac{2}{3} = 5^3 - 2^2 = 125 - 4 = 121$ 3) $8^\frac{9}{7} : 8^\frac{2}{7} - 3^\frac{6}{5} \cdot 3^{- \frac{4}{5}} = 8^{\frac{9}{7} - \frac{2}{7}} - 3^{\frac{6}{5} + (- \frac{4}{5})} = 8^{\frac{7}{7}} - 3^{\frac{2}{5}} = 8 - 3^{\frac{2}{5}}$ 4) $(5^{- \frac{2}{5}})^{-5} + (0.2)^4 = 5^{(- \frac{2}{5}) \cdot (-5)} + (\frac{1}{5})^{-4} = 5^2 + (5^{-1})^{-4} = 25 + 5^4 = 25 + 625 = 650$ **Задание 5** 1) $\sqrt[3]{a} \cdot \sqrt[6]{a} = a^{\frac{1}{3}} \cdot a^{\frac{1}{6}} = a^{\frac{1}{3} + \frac{1}{6}} = a^{\frac{2}{6} + \frac{1}{6}} = a^{\frac{3}{6}} = a^{\frac{1}{2}} = \sqrt{a} = \sqrt{0.09} = 0.3$ 2) $\sqrt{b} : \sqrt[6]{b} = b^{\frac{1}{2}} : b^{\frac{1}{6}} = b^{\frac{1}{2} - \frac{1}{6}} = b^{\frac{3}{6} - \frac{1}{6}} = b^{\frac{2}{6}} = b^{\frac{1}{3}} = \sqrt[3]{b} = \sqrt[3]{27} = 3$ 3) $\frac{\sqrt[3]{b} \cdot \sqrt{b^2}}{\sqrt[6]{b}} = \frac{b^{\frac{1}{3}} \cdot b^{\frac{2}{2}}}{b^{\frac{1}{6}}} = \frac{b^{\frac{1}{3}} \cdot b}{b^{\frac{1}{6}}} = b^{\frac{1}{3} + 1 - \frac{1}{6}} = b^{\frac{2}{6} + \frac{6}{6} - \frac{1}{6}} = b^{\frac{7}{6}} = b^{1 \frac{1}{6}} = b \cdot \sqrt[6]{b} = 1.3 \cdot \sqrt[6]{1.3}$ 4) $\sqrt[3]{a} \cdot \sqrt{a} \cdot \sqrt[12]{a^5} = a^{\frac{1}{3}} \cdot a^{\frac{1}{2}} \cdot a^{\frac{5}{12}} = a^{\frac{1}{3} + \frac{1}{2} + \frac{5}{12}} = a^{\frac{4}{12} + \frac{6}{12} + \frac{5}{12}} = a^{\frac{15}{12}} = a^{\frac{5}{4}} = a^{1 \frac{1}{4}} = a \cdot \sqrt[4]{a} = 2.7 \cdot \sqrt[4]{2.7}$ **Задание 6** 1) $a^\frac{1}{3} \cdot \sqrt{a} = a^\frac{1}{3} \cdot a^\frac{1}{2} = a^{\frac{1}{3} + \frac{1}{2}} = a^{\frac{2}{6} + \frac{3}{6}} = a^{\frac{5}{6}}$ 2) $b^2 \cdot b^\frac{2}{3} \cdot \sqrt[6]{b} = b^2 \cdot b^\frac{2}{3} \cdot b^\frac{1}{6} = b^{2 + \frac{2}{3} + \frac{1}{6}} = b^{\frac{12}{6} + \frac{4}{6} + \frac{1}{6}} = b^{\frac{17}{6}}$ 3) $\sqrt[3]{b} : b^6 = b^\frac{1}{3} : b^6 = b^{\frac{1}{3} - 6} = b^{\frac{1}{3} - \frac{18}{3}} = b^{- \frac{17}{3}}$ Надеюсь, теперь тебе стало понятнее, как решать такие примеры! Если что, спрашивай ещё!