Question

Решите срочно пожалуйста!!!!!

Answer 1

$\displaystyle \tt 1.\\\\a). \ \ 2^{-3}\cdot2^{-2}=2^{-5}=\frac{1}{2^{5}}=\frac{1}{32};\\\\b). \ \ \bigg(\frac{1}{8}\bigg)^{-10}:\bigg(\frac{1}{8}\bigg)^{-8}=8^{10}:8^{8}=8^{2}=64;\\\\c). \ \ (3^{-4})^{5}\cdot3^{16}=3^{-20}\cdot3^{16}=3^{-4}=\frac{1}{81};\\\\d). \ \ \bigg(1\frac{1}{9}\bigg)^{-2}=\bigg(\frac{10}{9}\bigg)^{-2}=\bigg(\frac{9}{10}\bigg)^{2}=0,81;$

$\displaystyle \tt 2.\\\\a). \ \ 16^{-3}\cdot8^{4}=(2^{4})^{-3}\cdot(2^{3})^{4}=2^{-12}\cdot2^{12}=2^{0}=1;\\\\b). \ \ (31^{-3})^{7}:31^{-20}=31^{-21}\cdot31^{20}=31^{-1}=\frac{1}{31};\\\\c). \ \ 81^{5}\cdot9^{-11}:3^{-4}=(3^{4})^{5}\cdot(3^{2})^{-11}\cdot3^{4}=3^{20}\cdot3^{-22}\cdot3^{4}=3^{2}=9;\\\\d). \ \ 125^{-7}:(5^{0}\cdot25^{-11})=(5^{3})^{-7}\cdot(5^{2})^{11}=5^{-21}\cdot5^{22}=5;$

$\displaystyle \tt 3. \ \ (6^{3-n})^{4}\cdot6^{2n-8}\cdot36=6^{12-4n}\cdot6^{2n-8}\cdot6^{2}=6^{6-2n};\\\\\\4.\\\\a). \ \ \frac{15x^{-6}}{y^{-5}}\cdot\frac{y^{-2}}{25x^{3}}=15x^{-6}\cdot y^{5}\cdot y^{-2}:25x^{3}=0,6x^{-9}y^{3};$

$\displaystyle \tt b). \ \ \bigg(\frac{9}{a^{8}b^{-5}}\bigg)^{-4}\cdot\bigg(\frac{a^{4}b^{-6}}{27}\bigg)^{-3}=\frac{a^{32}b^{-20}}{3^{8}}\cdot\frac{3^{9}}{a^{12}b^{-18}}=3a^{20}b^{-2};$