6) a) 81^-2 * 27^2; 6) 16^-5 : 8^-6; B) ((-6)^-9 * 6^-7) / 6^-15; r) (4^-6 * 16^-5) / 8^-10;

Решение:

1. \( 81^{-2} \cdot 27^2 = (3^4)^{-2} \cdot (3^3)^2 = 3^{-8} \cdot 3^6 = 3^{-8+6} = 3^{-2} = \frac{1}{9} \)
2. \( 16^{-5} : 8^{-6} = (2^4)^{-5} : (2^3)^{-6} = 2^{-20} : 2^{-18} = 2^{-20 - (-18)} = 2^{-20+18} = 2^{-2} = \frac{1}{4} \)
3. \( \frac{(-6)^{-9} \cdot 6^{-7}}{6^{-15}} = \frac{-(6^{-9}) \cdot 6^{-7}}{6^{-15}} = -\frac{6^{-9} \cdot 6^{-7}}{6^{-15}} = -\frac{6^{-9+(-7)}}{6^{-15}} = -\frac{6^{-16}}{6^{-15}} = -(6^{-16 - (-15)}) = -(6^{-16+15}) = -6^{-1} = -\frac{1}{6} \)
4. \( \frac{4^{-6} \cdot 16^{-5}}{8^{-10}} = \frac{(2^2)^{-6} \cdot (2^4)^{-5}}{(2^3)^{-10}} = \frac{2^{-12} \cdot 2^{-20}}{2^{-30}} = \frac{2^{-12+(-20)}}{2^{-30}} = \frac{2^{-32}}{2^{-30}} = 2^{-32 - (-30)} = 2^{-32+30} = 2^{-2} = \frac{1}{4} \)

Ответ: 1/9; 1/4; -1/6; 1/4.