Ответ:
Решение:
- \( \sqrt[3]{b} : b^{\frac{1}{6}} = b^{\frac{1}{3}} : b^{\frac{1}{6}} = b^{\frac{1}{3} - \frac{1}{6}} = b^{\frac{2-1}{6}} = b^{\frac{1}{6}} = \sqrt[6]{b} \)
- \( 81^{\frac{1}{4}} - (2\sqrt{3})^2 = (3^4)^{\frac{1}{4}} - (4 \cdot 3) = 3 - 12 = -9 \)
- \( \log_4 48 - \log_4 3 + 6 \log_6 5 = \log_4 \frac{48}{3} + \log_6 5^6 = \log_4 16 + \log_6 5^6 = 2 + \log_6 5^6 \)
- \( \frac{a^3 - 4}{a^3 - 2} - \frac{a^3}{a^3 - 2} = \frac{a^3 - 4 - a^3}{a^3 - 2} = \frac{-4}{a^3 - 2} \)
Ответ: а) \( \sqrt[6]{b} \); б) \( -9 \); в) \( 2 + \log_6 5^6 \); г) \( \frac{-4}{a^3 - 2} \).
