Решите.................

Решите задачу:

$1)\\ \\ 3*(( \frac{1}{6} )^{2+log_{ \frac{1}{6}}20}- \sqrt{2}^{4+log_{ \sqrt{2}} \frac{1}{9}})=\\ \\ 3*(( \frac{1}{6} )^{log_{ \frac{1}{6} } \frac{1}{36} +log_{ \frac{1}{6}}20}- \sqrt{2}^{log_{ \sqrt{2} }4+log_{ \sqrt{2}} \frac{1}{9}}) = \\ \\ 3*(( \frac{1}{6} )^{log_{ \frac{1}{6}} \frac{20}{36} }- \sqrt{2}^{log_{ \sqrt{2}} \frac{4}{9}}) = \\ \\ 3*( \frac{20}{36} - \frac{4}{9}) = 3*( \frac{20}{36} - \frac{16}{36}) = \\ \\ = 3*\frac{4}{36} = 3*\frac{1}{9} = \frac{1}{3}$
$2)\\\\ 25^{log_{5}3 - log_{125}2}-3^{-log_{3}8} = \\\\ = 25^{2log_{25}3-\frac{1}{2}log_{25}2} - \frac{1}{3^{log_{3}8}} = \\\\ = 5^{4log_{25}3-log_{25}2} - \frac{1}{3^{3log_{3}2}} = \\\\ = 5^{log_{25}3^{4}-log_{25}2} - \frac{1}{8}} = \\\\ = 5^{log_{25}\frac{3^{4}}{2}} - \frac{1}{8}} = \\\\ = 5^{\frac{1}{2}log_{5}\frac{3^{4}}{2}} - \frac{1}{8}} = \\\\ = \sqrt{\frac{3^{4}}{2}} - \frac{1}{8} = \\\\ = \frac{9}{\sqrt{2}} - \frac{1}{8} = \\\\$
$= \frac{9\sqrt{2}}{2} - \frac{1}{8} = \frac{36\sqrt{2}}{8}-\frac{1}{8} = \frac{36\sqrt{2}-1}{8}$