Решение

1. Вычислите:

a) $$16^{-\frac{3}{4}} = (2^4)^{-\frac{3}{4}} = 2^{4 \cdot (-\frac{3}{4})} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$

б) $$(\frac{1}{125})^{-\frac{2}{3}} = (5^{-3})^{-\frac{2}{3}} = 5^{-3 \cdot (-\frac{2}{3})} = 5^2 = 25$$

2. Упростите:

a) $$(\sqrt{a} + \sqrt{b})(\sqrt[4]{a} + \sqrt[4]{b})(\sqrt[4]{a} - \sqrt[4]{b}) = (\sqrt{a} + \sqrt{b})((\sqrt[4]{a})^2 - (\sqrt[4]{b})^2) = (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a - b$$

б) $$\frac{a^{\frac{3}{4}} - a^{-\frac{1}{4}}}{a - a^{\frac{1}{2}}} (a^{\frac{1}{4}} + 1) = \frac{a^{-\frac{1}{4}}(a - 1)}{a^{\frac{1}{2}}(a^{\frac{1}{2}} - 1)} (a^{\frac{1}{4}} + 1) = \frac{a^{-\frac{1}{4}}(a^{\frac{1}{2}} + 1)(a^{\frac{1}{2}} - 1)}{a^{\frac{1}{2}}(a^{\frac{1}{2}} - 1)} (a^{\frac{1}{4}} + 1) = \frac{a^{-\frac{1}{4}} (a^{\frac{1}{2}} + 1)}{a^{\frac{1}{2}}} (a^{\frac{1}{4}} + 1) = a^{-\frac{1}{4} - \frac{1}{2}} (a^{\frac{1}{2}} + 1) (a^{\frac{1}{4}} + 1) = a^{-\frac{3}{4}} (a^{\frac{1}{2}} + 1) (a^{\frac{1}{4}} + 1)$$ $$= a^{-\frac{3}{4}} (a^{\frac{3}{4}} + a^{\frac{1}{2}} + a^{\frac{1}{4}} + 1) = a^{-\frac{3}{4}} a^{\frac{3}{4}} + a^{-\frac{3}{4}} a^{\frac{1}{2}} + a^{-\frac{3}{4}} a^{\frac{1}{4}} + a^{-\frac{3}{4}} = 1 + a^{-\frac{1}{4}} + a^{-\frac{1}{2}} + a^{-\frac{3}{4}} = 1 + \frac{1}{\sqrt[4]{a}} + \frac{1}{\sqrt{a}} + \frac{1}{\sqrt[4]{a^3}}$$

3. Упростите и вычислите

$$\left(\frac{9a^{-\frac{5}{24}}}{a^{\frac{8}{3}} a^{\frac{1}{3}}}\right)^{\frac{1}{3}} = \left(\frac{9a^{-\frac{5}{24}}}{a^{\frac{9}{3}}}\right)^{\frac{1}{3}} = \left(\frac{9a^{-\frac{5}{24}}}{a^{3}}\right)^{\frac{1}{3}} = \left(9a^{-\frac{5}{24} - 3}\right)^{\frac{1}{3}} = \left(9a^{-\frac{5+72}{24}} \right)^{\frac{1}{3}} = \left(9a^{-\frac{77}{24}}\right)^{\frac{1}{3}} = 9^{\frac{1}{3}} a^{-\frac{77}{24} \cdot \frac{1}{3}} = \sqrt[3]{9} a^{-\frac{77}{72}} = \sqrt[3]{9} \cdot (24)^{-\frac{77}{72}} = \sqrt[3]{9} \cdot (2^3 \cdot 3)^{-\frac{77}{72}} = \sqrt[3]{9} \cdot 2^{-\frac{77}{24}} \cdot 3^{-\frac{77}{72}} $$

При $$a = 24$$:

$$\sqrt[3]{9} \cdot (24)^{-\frac{77}{72}} = \sqrt[3]{9} \cdot 2^{-\frac{77}{24}} \cdot 3^{-\frac{77}{72}} $$