Ответ:

Решение:

Воспользуемся правилом \( a^{-n} = \frac{1}{a^n} \) и \( (\frac{a}{b})^{-n} = (\frac{b}{a})^n \).

  1. \( 4^{-2} = \frac{1}{4^2} = \frac{1}{16} \)
  2. \( (-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{-27} = -\frac{1}{27} \)
  3. \( (-1)^{-9} = \frac{1}{(-1)^9} = \frac{1}{-1} = -1 \)
  4. \( \left(\frac{1}{7}\right)^{-2} = \left(\frac{7}{1}\right)^2 = 7^2 = 49 \)
  5. \( \left(1\frac{1}{2}\right)^{-5} = \left(\frac{3}{2}\right)^{-5} = \left(\frac{2}{3}\right)^5 = \frac{2^5}{3^5} = \frac{32}{243} \)
  6. \( \left(-\frac{2}{3}\right)^{-3} = \left(-\frac{3}{2}\right)^3 = (-\frac{3^3}{2^3}) = -\frac{27}{8} \)
  7. \( (-1)^{-20} = \frac{1}{(-1)^{20}} = \frac{1}{1} = 1 \)
  8. \( \left(-2\frac{2}{5}\right)^{-2} = \left(-\frac{12}{5}\right)^{-2} = \left(-\frac{5}{12}\right)^2 = \frac{(-5)^2}{12^2} = \frac{25}{144} \)
  9. \( 0,01 = \frac{1}{100} \)
  10. \( 1,1 = \frac{11}{10} \)

Ответ: а) \(\frac{1}{16}\); б) \(-\frac{1}{27}\); в) \(-1\); г) \(49\); д) \(\frac{32}{243}\); е) \(-\frac{27}{8}\); ж) \(1\); з) \(\frac{25}{144}\); и) \(\frac{1}{100}\); к) \(\frac{11}{10}\).