a) (7-4-(5/7))^0 - 2 б) (3(2/3)^-1 + 4^-1) / (2.9^0 - 0.1^-1) в) ((4/3)^-1 (2/3)^-1) (6/5)^-2 - 6.9^0 : (-1/11)^-1 e) (3/8)^-1 * 1.5^3 + (3/8)^-1 * 1.5^3 e) (0.5^-2 - 5*(-2)^-2 + (2/3)^2) / (2^-2 + 12.7^0)

Решение:

Применяем свойства степени: \( a^0 = 1 \), \( a^{-n} = \frac{1}{a^n} \), \( (a/b)^{-n} = (b/a)^n \), \( a^m \cdot a^n = a^{m+n} \), \( a^m : a^n = a^{m-n} \), \( (a^m)^n = a^{m · n} \).

1. а) \( (7 - 4 - (5/7))^0 - 2 = (3 - 5/7)^0 - 2 = (21/7 - 5/7)^0 - 2 = (16/7)^0 - 2 = 1 - 2 = -1 \)
2. б) \( \frac{3 \cdot (2/3)^{-1} + 4^{-1}}{2.9^0 - 0.1^{-1}} = \frac{3 \cdot (3/2) + 1/4}{1 - 10} = \frac{9/2 + 1/4}{ -9} = \frac{18/4 + 1/4}{-9} = \frac{19/4}{-9} = -\frac{19}{36} \)
3. в) \( ((\frac{4}{3})^{-1} \cdot (\frac{2}{3})^{-1}) \cdot (\frac{6}{5})^{-2} - 6.9^0 : (-\frac{1}{11})^{-1} = (\frac{3}{4} \cdot \frac{3}{2}) \cdot (\frac{5}{6})^2 - 1 : (-11) = \frac{9}{8} \cdot \frac{25}{36} - (-11) = \frac{1}{8} \cdot \frac{25}{4} + 11 = \frac{25}{32} + 11 = \frac{25 + 352}{32} = \frac{377}{32} \)
4. г) \( (\frac{3}{8})^{-1} \cdot 1.5^3 + (\frac{3}{8})^{-1} \cdot 1.5^3 = \frac{8}{3} \cdot (\frac{3}{2})^3 + \frac{8}{3} \cdot (\frac{3}{2})^3 = \frac{8}{3} \cdot \frac{27}{8} + \frac{8}{3} \cdot \frac{27}{8} = 9 + 9 = 18 \)
5. д) \( \frac{0.5^{-2} - 5 \cdot (-2)^{-2} + (2/3)^2}{2^{-2} + 12.7^0} = \frac{(1/2)^{-2} - 5 \cdot (1/(-2))^2 + 4/9}{1/4 + 1} = \frac{2^2 - 5 \cdot (1/4) + 4/9}{5/4} = \frac{4 - 5/4 + 4/9}{5/4} = \frac{(144 - 45 + 16)/36}{5/4} = \frac{115/36}{5/4} = \frac{115}{36} \cdot \frac{4}{5} = \frac{23}{9} \)

Ответ: а) -1; б) -\(\frac{19}{36}\); в) \(\frac{377}{32}\); г) 18; д) \(\frac{23}{9}\).