a)
[tex]\it \dfrac{6^{-10}}{81^{-2}\cdot32^{-2}}= \dfrac{(2\cdot3)^{-10}}{(3^4)^{-2}\cdot(2^5)^{-2}} =\dfrac{2^{-10}\cdot 3^{-10}}{3^{-8}\cdot2^{-10}} =\dfrac{3^{-10}}{3^{-8}} =
\\\;\\ \\\;\\
=\dfrac{3^{-8}\cdot3^{-2}}{3^{-8}} = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}[/tex]
