2. Найдите значение выражения: а) \frac{\sqrt{81\sqrt{b}}}{\sqrt[14]{b}} при b > 0. б) \frac{(\sqrt[3]{7a^2})^6}{a^4} при а ≠ 0. 25.24 В) ( \frac{2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}}}{\sqrt[12]{2}})^2\cdot г) \frac{9^{\log_5{50}}}{9^{\log_5{2}}}.

Решение:

а) \( \frac{\sqrt{81\sqrt{b}}}{\sqrt[14]{b}} = \frac{\sqrt{81} \cdot \sqrt{\sqrt{b}}}{b^{\frac{1}{14}}} = \frac{9 \cdot b^{\frac{1}{4}}}{b^{\frac{1}{14}}} = 9 \cdot b^{\frac{1}{4} - \frac{1}{14}} = 9 \cdot b^{\frac{7}{28} - \frac{2}{28}} = 9 \cdot b^{\frac{5}{28}} = 9 \sqrt[28]{b^5} \)

б) \( \frac{(\sqrt[3]{7a^2})^6}{a^4} = \frac{(7a^2)^{\frac{6}{3}}}{a^4} = \frac{(7a^2)^2}{a^4} = \frac{49a^4}{a^4} = 49 \)

в) \( ( \frac{2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}}}{\sqrt[12]{2}})^2 = ( \frac{2^{\frac{1}{3} + \frac{1}{4}}}{2^{\frac{1}{12}}})^2 = ( \frac{2^{\frac{7}{12}}}{2^{\frac{1}{12}}})^2 = (2^{\frac{7}{12} - \frac{1}{12}})^2 = (2^{\frac{6}{12}})^2 = (2^{\frac{1}{2}})^2 = 2 \)

г) \( \frac{9^{\log_5{50}}}{9^{\log_5{2}}} = 9^{\log_5{50} - \log_5{2}} = 9^{\log_5{\frac{50}{2}}} = 9^{\log_5{25}} = 9^2 = 81 \)

Ответ: а) \( 9 \sqrt[28]{b^5} \), б) \( 49 \), в) \( 2 \), г) \( 81 \)