Задание 577

Решение:

• \({\largeа)}\ 4^{{-}2}=\frac{1}{4^2}=\frac{1}{4\cdot4}=\frac{1}{16}\)

• \({\largeб)}\ 3^{{-}1}=\frac{1}{3^1}=\frac{1}{3}\)

• \({\largeв)}\ 3^{{-}4}=\frac{1}{3^4}=\frac{1}{3\cdot3\cdot3\cdot3}=\frac{1}{81}\)

• \({\largeг)}\ 7{,}12^0=1\)

• \({\largeд)}\ 5^{{-}1}+4^{{-}1}=\frac{1}{5^1}+\frac{1}{4^1}=\frac{1^{\backslash4}}{5\phantom{^{\backslash4}}}+\frac{1^{\backslash5}}{4\phantom{^{\backslash5}}}=\frac{1\cdot4}{5\cdot4}+\frac{1\cdot5}{4\cdot5}=\frac{4}{20}+\frac{5}{20}=\frac{9}{20}\)

• \({\largeе)}\ (5+4)^{{-}1}=9^{{-1}}=\frac{1}{9^1}=\frac{1}{9}\)

• \({\largeж)}\ 4^{{-}1}-5^{{-}1}=\frac{1}{4^1}-\frac{1}{5^1}=\frac{1^{\backslash5}}{4\phantom{^{\backslash5}}}-\frac{1^{\backslash4}}{5\phantom{^{\backslash4}}}=\frac{1\cdot5}{4\cdot5}-\frac{1\cdot4}{5\cdot4}=\frac{5}{20}-\frac{4}{20}=\frac{1}{20}\)

• \({\largeз)}\ (3^{{-}1}-5^{{-}1})^{{-}2}=\left(\frac{1}{3^1}-\frac{1}{5^1}\right)^{{-}2}=\left(\frac{1^{\backslash5}}{3\phantom{^{\backslash5}}}-\frac{1^{\backslash3}}{5\phantom{^{\backslash3}}}\right)^{{-}2}=\left(\frac{1\cdot5}{3\cdot5}-\frac{1\cdot3}{5\cdot3}\right)^{{-}2}=\left(\frac{5}{15}-\frac{3}{15}\right)^{{-}2}=\left(\frac{2}{15}\right)^{{-}2}=\left(\frac{15}{2}\right)^2=\frac{15\cdot15}{2\cdot2}=\frac{225}{4}=56\frac{1}{4}\)

• \({\largeи)}\ 2^{{-}3}+4^{{-}2}=\frac{1}{2^3}+\frac{1}{4^2}=\frac{1}{2\cdot2\cdot2}+\frac{1}{4\cdot4}=\frac{1^{\backslash2}}{8\phantom{^{\backslash2}}}+\frac{1^{\backslash1}}{16\phantom{^{\backslash1}}}=\frac{1\cdot2}{8\cdot2}+\frac{1\cdot1}{16\cdot1}=\frac{2}{16}+\frac{1}{16}=\frac{3}{16}\)

• \({\largeк)}\ 3^{{-}2}-9^{{-}1}=\frac{1}{3^2}-\frac{1}{9^1}=\frac{1}{3\cdot3}-\frac{1}{9}=\frac{1}{9}-\frac{1}{9}=0\)

• \({\largeл)}\ 4^2\cdot2^{{-}3}=4^2\cdot\frac{1}{2^3}=4\cdot4\cdot\frac{1}{2\cdot2\cdot2}=16\cdot\frac{1}{8}=2\)

• \({\largeм)}\ 3^{{-}4}:9^{{-}2}=\frac{1}{3^4}:\frac{1}{9^2}=\frac{1}{3\cdot3\cdot3\cdot3}:\frac{1}{9\cdot9}=\frac{1}{81}:\frac{1}{81}=\frac{1}{81}\cdot\frac{81}{1}=1\)