7) 6-10 81⁻² ⋅ 16⁻³, 14⁵ ⋅ 2⁻⁷

7) $$6^{-10}\cdot\frac{81^{-2}\cdot16^{-3}}{14^5\cdot2^{-7}} = (2\cdot3)^{-10}\cdot\frac{(3^4)^{-2}\cdot(2^4)^{-3}}{(2\cdot7)^5\cdot2^{-7}} = 2^{-10}\cdot3^{-10}\cdot\frac{3^{-8}\cdot2^{-12}}{2^5\cdot7^5\cdot2^{-7}} = \frac{2^{-10}\cdot3^{-10}\cdot3^{-8}\cdot2^{-12}}{2^5\cdot7^5\cdot2^{-7}} = \frac{2^{-10-12}\cdot3^{-10-8}}{2^{5-7}\cdot7^5} = \frac{2^{-22}\cdot3^{-18}}{2^{-2}\cdot7^5} = 2^{-22-(-2)}\cdot3^{-18}\cdot7^{-5} = 2^{-20}\cdot3^{-18}\cdot7^{-5} = \frac{1}{2^{20}\cdot3^{18}\cdot7^5}$$.

Ответ: $$\frac{1}{2^{20}\cdot3^{18}\cdot7^5}$$