Вычислите: 1) (64 \cdot 4⁻⁷)² \cdot (16⁻¹)⁻³; 2) 81⁻⁸ \cdot 27⁻⁵ / 9⁻¹²

Ответ:

• 1) \((64 \cdot 4^{-7})^{2} \cdot (16^{-1})^{-3} = (4^{3} \cdot 4^{-7})^{2} \cdot 16^{3} = (4^{3-7})^{2} \cdot (4^{2})^{3} = (4^{-4})^{2} \cdot 4^{6} = 4^{-8} \cdot 4^{6} = 4^{-8+6} = 4^{-2} = (\frac{1}{4})^{2} = \frac{1}{16}\)

• 2) \(\frac{81^{-8} \cdot 27^{-5}}{9^{-12}} = \frac{(3^{4})^{-8} \cdot (3^{3})^{-5}}{(3^{2})^{-12}} = \frac{3^{-32} \cdot 3^{-15}}{3^{-24}} = \frac{3^{-32-15}}{3^{-24}} = \frac{3^{-47}}{3^{-24}} = 3^{-47-(-24)} = 3^{-47+24} = 3^{-23} = (\frac{1}{3})^{23}\)