Используем правило: \(a^{-n}=\frac{1}{a^n}\), где \(a
e0\).
- \(12^{-2}=\frac{1}{12^2}=\frac{1}{144}\);
- \(3^{-4}=\frac{1}{3^4}=\frac{1}{81}\);
- \((-2)^{-6}=\frac{1}{(-2)^6}=\frac{1}{64}\);
- \((-5)^{-3}=\frac{1}{(-5)^3}=-\frac{1}{125}\);
- \(\left(-\frac{1}{8}\right)^{-1}=-8\);
- \(\left(\frac{2}{3}\right)^{-3}=\left(\frac{3}{2}\right)^3=\frac{27}{8}\);
- \(\left(-\frac{7}{9}\right)^{-2}=\left(-\frac{9}{7}\right)^2=\frac{81}{49}\);
- \(\left(1\frac{2}{3}\right)^{-1}=\left(\frac{5}{3}\right)^{-1}=\frac{3}{5}\);
- \(0.3^{-2}=\left(\frac{3}{10}\right)^{-2}=\left(\frac{10}{3}\right)^2=\frac{100}{9}\);
- \(1.6^{-2}=\left(\frac{8}{5}\right)^{-2}=\left(\frac{5}{8}\right)^2=\frac{25}{64}\).
Ответ: 1) \(\frac{1}{144}\); 2) \(\frac{1}{81}\); 3) \(\frac{1}{64}\); 4) \(-\frac{1}{125}\); 5) \(-8\); 6) \(\frac{27}{8}\); 7) \(\frac{81}{49}\); 8) \(\frac{3}{5}\); 9) \(\frac{100}{9}\); 10) \(\frac{25}{64}\).