1. Вычислите: a) 27^{-3} \(\cdot\) 81^2; 6) \(\frac{14^2 \cdot (28^{-2})^3}{(98^{-1})^2}\); B) \(\frac\){2^{-2} - \(\frac{2}{3}\)^{-1}}{2 - (-1,25)^2} \(\cdot\) \(3^0 - \frac{1}{7}\)^{-1} + 15 \(\cdot\) 3^{-2}; r) (0,3)^{-2} \(\cdot\) (0,09)^{-1} \(\cdot\) \(-3\frac{1}{3}\)^{-3}; (\(-\frac{\sqrt{5}}{2}\)^{-2} + \(-\frac{1}{3}\)^{-1} + \(5\frac{2}{3}\)^0)

Решение:

1. \( 27^{-3} \cdot 81^2 = (3^3)^{-3} \cdot (3^4)^2 = 3^{-9} \cdot 3^8 = 3^{-1} = \frac{1}{3} \)
2. \( \frac{14^2 \cdot (28^{-2})^3}{(98^{-1})^2} = \frac{(2 \cdot 7)^2 \cdot (2^2 \cdot 7)^{-6}}{(2 \cdot 7^2)^{-2}} = \frac{2^2 \cdot 7^2 \cdot 2^{-12} \cdot 7^{-6}}{2^{-2} \cdot 7^{-4}} = 2^{2-12-(-2)} \cdot 7^{2-6-(-4)} = 2^{-8} \cdot 7^0 = \frac{1}{2^8} = \frac{1}{256} \)
3. \( \frac{2^{-2} - (\frac{2}{3})^{-1}}{2 - (-1,25)^2} \cdot (3^0 - \frac{1}{7})^{-1} + 15 \cdot 3^{-2} = \frac{\frac{1}{4} - \frac{3}{2}}{2 - (\frac{5}{4})^2} \cdot (1 - \frac{1}{7})^{-1} + 15 \cdot \frac{1}{9} = \frac{\frac{1-6}{4}}{2 - \frac{25}{16}} \cdot (\frac{6}{7})^{-1} + \frac{5}{3} = \frac{-\frac{5}{4}}{\frac{32-25}{16}} \cdot \frac{7}{6} + \frac{5}{3} = \frac{-\frac{5}{4}}{\frac{7}{16}} \cdot \frac{7}{6} + \frac{5}{3} = \frac{-5}{4} \cdot \frac{16}{7} \cdot \frac{7}{6} + \frac{5}{3} = \frac{-5 \cdot 4}{6} + \frac{5}{3} = \frac{-20}{6} + \frac{5}{3} = \frac{-10}{3} + \frac{5}{3} = \frac{-5}{3} \)
4. \( (0,3)^{-2} \cdot (0,09)^{-1} \cdot (-3\frac{1}{3})^{-3} = (\frac{3}{10})^{-2} \cdot (\frac{9}{100})^{-1} \cdot (-\frac{10}{3})^{-3} = (\frac{10}{3})^2 \cdot (\frac{100}{9}) \cdot (-\frac{3}{10})^3 = \frac{100}{9} \cdot \frac{100}{9} \cdot (-\frac{27}{1000}) = \frac{100}{9} \cdot \frac{100}{9} \cdot (-\frac{27}{1000}) = \frac{10000}{81} \cdot (-\frac{27}{1000}) = \frac{10}{1} \cdot (-\frac{1}{3}) = -\frac{10}{3} \)
5. \( (-\frac{\sqrt{5}}{2})^{-2} + (-\frac{1}{3})^{-1} + (5\frac{2}{3})^0 = (\frac{2}{-\sqrt{5}})^2 + (-3) + 1 = (\frac{4}{5}) - 3 + 1 = \frac{4}{5} - 2 = \frac{4-10}{5} = -\frac{6}{5} \)

Ответ: a) 1/3; б) 1/256; в) -5/3; г) -10/3; д) -6/5.