Решение №6

a) $$(27\cdot 3^{-6})^2\cdot (9^{-1})^{-2} = (3^3\cdot 3^{-6})^2\cdot (3^{-2})^{-2} = (3^{3-6})^2\cdot 3^{(-2)\cdot(-2)} = (3^{-3})^2\cdot 3^{4} = 3^{-6}\cdot 3^{4} = 3^{-6+4} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$$

б) $$\frac{64^{-4}\cdot 8^{3}}{16^{-3}} = \frac{(2^6)^{-4}\cdot (2^3)^{3}}{(2^4)^{-3}} = \frac{2^{-24}\cdot 2^9}{2^{-12}} = \frac{2^{-15}}{2^{-12}} = 2^{-15-(-12)} = 2^{-15+12} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$$