Ответ:
Используем правило: \(a^{-n}=\frac{1}{a^n}\).
- \(4^{-2}=\frac{1}{4^2}=\frac{1}{16}\).
- \((-3)^{-3}=\frac{1}{(-3)^3}=-\frac{1}{27}\).
- \((-1)^{-9}=\frac{1}{(-1)^9}=-1\).
- \((-1)^{-20}=\frac{1}{(-1)^{20}}=1\).
- \(\left(\frac{1}{7}\right)^{-2}=7^2=49\).
- \(\left(-\frac{2}{3}\right)^{-3}=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}\).
- \(\left(1\frac{1}{2}\right)^{-5}=\left(\frac{3}{2}\right)^{-5}=\left(\frac{2}{3}\right)^5=\frac{32}{243}\).
- \(\left(-2\frac{2}{5}\right)^{-2}=\left(-\frac{12}{5}\right)^{-2}=\left(-\frac{5}{12}\right)^2=\frac{25}{144}\).
- \(0{,}01^{-2}=\left(\frac{1}{100}\right)^{-2}=100^2=10000\).
- \(1{,}125^{-1}=\left(\frac{9}{8}\right)^{-1}=\frac{8}{9}\).
Ответ: а) \(\frac{1}{16}\); б) \(-\frac{1}{27}\); в) \(-1\); г) \(1\); д) \(49\); е) \(-\frac{27}{8}\); ж) \(\frac{32}{243}\); з) \(\frac{25}{144}\); и) \(10000\); к) \(\frac{8}{9}\).
