4. Вычислим:
- a) $$\frac{4^{-8} \cdot 4^{-10}}{4^{-16}} = \frac{4^{-18}}{4^{-16}} = 4^{-18 - (-16)} = 4^{-18 + 16} = 4^{-2} = \frac{1}{4^2} = \frac{1}{16}$$.
- б) $$6^{-1} + 3^{-2} = \frac{1}{6} + \frac{1}{3^2} = \frac{1}{6} + \frac{1}{9} = \frac{3}{18} + \frac{2}{18} = \frac{5}{18}$$.
- в) $$(27 \cdot 3^{-7})^2 \cdot (9^{-1})^2 = (3^3 \cdot 3^{-7})^2 \cdot (3^{-2})^2 = (3^{-4})^2 \cdot 3^{-4} = 3^{-8} \cdot 3^{-4} = 3^{-12} = \frac{1}{3^{12}} = \frac{1}{531441}$$.
- г) $$\frac{4^{-4} \cdot 8^{-3}}{32^{-3}} = \frac{4^{-4} \cdot (2^3)^{-3}}{(2^5)^{-3}} = \frac{(2^2)^{-4} \cdot 2^{-9}}{2^{-15}} = \frac{2^{-8} \cdot 2^{-9}}{2^{-15}} = \frac{2^{-17}}{2^{-15}} = 2^{-17 - (-15)} = 2^{-17 + 15} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}$$.
Ответ: a) $$\frac{1}{16}$$; б) $$\frac{5}{18}$$; в) $$\frac{1}{531441}$$; г) $$\frac{1}{4}$$.