1. $6 \cdot (\frac{1}{3})^2 - 11 \frac{1}{3} = 6 \cdot \frac{1}{9} - \frac{34}{3} = \frac{2}{3} - \frac{34}{3} = -\frac{32}{3} = -10 \frac{2}{3}$
2. $9.3 + 7.8 = 17.1$
3. $(6.7 \cdot 10^{-3}) \cdot (5 \cdot 10^{-3}) = 6.7 \cdot 5 \cdot 10^{-3} \cdot 10^{-3} = 33.5 \cdot 10^{-6} = 3.35 \cdot 10^{-5}$
4. $\frac{0.9}{1 + \frac{1}{8}} = \frac{0.9}{\frac{9}{8}} = 0.9 \cdot \frac{8}{9} = 0.1 \cdot 8 = 0.8$
5. $(\frac{7}{18} + \frac{13}{20}) : \frac{17}{36} = (\frac{7 \cdot 10}{18 \cdot 10} + \frac{13 \cdot 9}{20 \cdot 9}) : \frac{17}{36} = (\frac{70}{180} + \frac{117}{180}) : \frac{17}{36} = \frac{187}{180} : \frac{17}{36} = \frac{187}{180} \cdot \frac{36}{17} = \frac{187}{5} \cdot \frac{1}{17} = \frac{11}{5} = 2 \frac{1}{5} = 2.2$
6. $\frac{2.7}{2.9 - 1.1} = \frac{2.7}{1.8} = \frac{27}{18} = \frac{3}{2} = 1.5$
7. $0.6 \cdot (-10)^4 + 4 \cdot (-10)^3 + 70 = 0.6 \cdot 10000 + 4 \cdot (-1000) + 70 = 6000 - 4000 + 70 = 2070$
8. $\frac{24}{3.2 \cdot 2} = \frac{24}{6.4} = \frac{240}{64} = \frac{120}{32} = \frac{60}{16} = \frac{30}{8} = \frac{15}{4} = 3.75$
9. $0.6 \cdot (-10)^3 + 50 = 0.6 \cdot (-1000) + 50 = -600 + 50 = -550$
10. $\frac{3}{4} + \frac{4}{5} = \frac{3 \cdot 5}{4 \cdot 5} + \frac{4 \cdot 4}{5 \cdot 4} = \frac{15}{20} + \frac{16}{20} = \frac{31}{20} = 1 \frac{11}{20} = 1.55$
11. $\frac{\sqrt{35} \cdot \sqrt{21}}{\sqrt{15}} = \frac{\sqrt{7 \cdot 5} \cdot \sqrt{7 \cdot 3}}{\sqrt{3 \cdot 5}} = \frac{\sqrt{7} \cdot \sqrt{5} \cdot \sqrt{7} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{5}} = \sqrt{7} \cdot \sqrt{7} = 7$
12. $\frac{(a-2b)^2 - 4b^2}{a} = \frac{a^2 - 4ab + 4b^2 - 4b^2}{a} = \frac{a^2 - 4ab}{a} = \frac{a(a - 4b)}{a} = a - 4b = 0.3 - 4 \cdot (-0.35) = 0.3 + 1.4 = 1.7$
13. $(8b-8)(8b+8) - 8b(8b+8) = 64b^2 - 64 - (64b^2 + 64b) = 64b^2 - 64 - 64b^2 - 64b = -64 - 64b = -64 - 64 \cdot 2.6 = -64 - 166.4 = -230.4$
14. $\frac{(a^3)^4}{a^9} = \frac{a^{12}}{a^9} = a^{12-9} = a^3 = 3^3 = 27$
15. $10ab + (-5a+b)^2 = 10ab + 25a^2 - 10ab + b^2 = 25a^2 + b^2 = 25 \cdot (\sqrt{10})^2 + (\sqrt{5})^2 = 25 \cdot 10 + 5 = 250 + 5 = 255$
16. $\frac{x^2 - 4}{4x^2} : \frac{2x}{x+2} = \frac{(x-2)(x+2)}{4x^2} \cdot \frac{x+2}{2x} = \frac{(x-2)(x+2)^2}{8x^3} = \frac{(4-2)(4+2)^2}{8 \cdot 4^3} = \frac{2 \cdot 36}{8 \cdot 64} = \frac{72}{512} = \frac{36}{256} = \frac{18}{128} = \frac{9}{64}$
17. $\frac{4ac^2}{a^2 - c^2} - \frac{a+c}{ac} = \frac{4 \cdot 3.1 \cdot 3.6^2}{3.1^2 - 3.6^2} - \frac{3.1+3.6}{3.1 \cdot 3.6} = \frac{4 \cdot 3.1 \cdot 12.96}{9.61 - 12.96} - \frac{6.7}{11.16} = \frac{160.704}{-3.35} - \frac{6.7}{11.16} = -47.97 - 0.599 = -48.569 \approx -48.57$
18. $a(a+1) - (a-3)^2 = a^2 + a - (a^2 - 6a + 9) = a^2 + a - a^2 + 6a - 9 = 7a - 9 = 7 \cdot (-1) - 9 = -7 - 9 = -16$
19. Допущение: $\sqrt{x} - 5\sqrt{y}$ в знаменателе. $\frac{4x - 25y}{2\sqrt{x} - 5\sqrt{y}} - 3\sqrt{y} = \frac{(2\sqrt{x} - 5\sqrt{y})(2\sqrt{x} + 5\sqrt{y})}{2\sqrt{x} - 5\sqrt{y}} - 3\sqrt{y} = 2\sqrt{x} + 5\sqrt{y} - 3\sqrt{y} = 2\sqrt{x} + 2\sqrt{y} = 2(\sqrt{x} + \sqrt{y}) = 2 \cdot 4 = 8$
20. $\frac{9b}{a-b} \cdot \frac{a^2 - ab}{54b} = \frac{9b}{a-b} \cdot \frac{a(a-b)}{54b} = \frac{9b \cdot a (a-b)}{(a-b) \cdot 54b} = \frac{9ab}{54b} = \frac{a}{6} = \frac{-63}{6} = -10.5$
21. $\frac{5ab}{5ab - 8a^2} = \frac{5 \cdot 3 \cdot 8}{5 \cdot 3 \cdot 8 - 8 \cdot 3^2} = \frac{120}{120 - 72} = \frac{120}{48} = \frac{60}{24} = \frac{30}{12} = \frac{15}{6} = \frac{5}{2} = 2.5$
22. $(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) = (\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2$
23. $(2 + \sqrt{3})^2 + (2 - \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 + 4 - 4\sqrt{3} + 3 = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14$
24. Допущение: в скобках должно быть $\frac{a}{8b} - \frac{8b}{a}$ . Тогда:
$\frac{8ab}{a+8b} \cdot (\frac{a}{8b} - \frac{8b}{a}) = \frac{8ab}{a+8b} \cdot (\frac{a^2 - 64b^2}{8ab}) = \frac{8ab \cdot (a^2 - 64b^2)}{(a+8b) \cdot 8ab} = \frac{a^2 - 64b^2}{a+8b} = \frac{(a - 8b)(a+8b)}{a+8b} = a - 8b = 8\sqrt{3} + 7 - 8(\sqrt{3} - 3) = 8\sqrt{3} + 7 - 8\sqrt{3} + 24 = 31$
25. $f(x-4) = 68 - x$. Пусть $x-4 = 2$, тогда $x = 6$. $f(2) = 68 - 6 = 62$
