6) a) 81⁻² ⋅ 27²; г) (4⁻⁶ ⋅ 16⁻⁵) / 8⁻¹⁰;

Решение:

1. \( 81^{-2} \cdot 27^2 = (3^4)^{-2} \cdot (3^3)^2 = 3^{4 \cdot (-2)} \cdot 3^{3 \cdot 2} = 3^{-8} \cdot 3^6 = 3^{-8+6} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \)
2. \( \frac{4^{-6} \cdot 16^{-5}}{8^{-10}} = \frac{(2^2)^{-6} \cdot (2^4)^{-5}}{(2^3)^{-10}} = \frac{2^{2 \cdot (-6)} \cdot 2^{4 \cdot (-5)}}{2^{3 \cdot (-10)}} = \frac{2^{-12} \cdot 2^{-20}}{2^{-30}} = \frac{2^{-12-20}}{2^{-30}} = \frac{2^{-32}}{2^{-30}} = 2^{-32 - (-30)} = 2^{-32+30} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \)

Ответ: 1) \(\frac{1}{9}\); 2) \(\frac{1}{4}\).