Решим данные выражения.
a) $$(\frac{4}{2} + \frac{9}{6}) \cdot \frac{9}{13} = (\frac{4}{2} + \frac{3}{2}) \cdot \frac{9}{13} = \frac{7}{2} \cdot \frac{9}{13} = \frac{7 \cdot 9}{2 \cdot 13} = \frac{63}{26} = 2 \frac{11}{26}$$.
б) $$(\frac{9}{4} - \frac{11}{11}) \cdot \frac{11}{5} = (\frac{9}{4} - 1) \cdot \frac{11}{5} = (\frac{9}{4} - \frac{4}{4}) \cdot \frac{11}{5} = \frac{5}{4} \cdot \frac{11}{5} = \frac{5 \cdot 11}{4 \cdot 5} = \frac{11}{4} = 2 \frac{3}{4}$$.
д) $$(2 \frac{3}{2})^2 + \frac{13}{21} \cdot \frac{26}{7} - \frac{18}{5} = (2 + \frac{3}{2})^2 + \frac{13}{21} \cdot \frac{26}{7} - \frac{18}{5} = (\frac{4}{2} + \frac{3}{2})^2 + \frac{13}{21} \cdot \frac{26}{7} - \frac{18}{5} = (\frac{7}{2})^2 + \frac{13 \cdot 26}{21 \cdot 7} - \frac{18}{5} = \frac{49}{4} + \frac{13 \cdot 13 \cdot 2}{7 \cdot 3 \cdot 7} - \frac{18}{5} = \frac{49}{4} + \frac{169 \cdot 2}{49 \cdot 3} - \frac{18}{5} = \frac{49}{4} + \frac{338}{147} - \frac{18}{5} = \frac{49 \cdot 147 \cdot 5 + 338 \cdot 4 \cdot 5 - 18 \cdot 4 \cdot 147}{4 \cdot 147 \cdot 5} = \frac{36015 + 6760 - 10584}{2940} = \frac{32191}{2940} = 10 \frac{2791}{2940}$$.
е) $$(3 \frac{1}{2})^2 \cdot (\frac{7}{3} - \frac{7}{3})^3 = (3 + \frac{1}{2})^2 \cdot (\frac{7}{3} - \frac{7}{3})^3 = (\frac{6}{2} + \frac{1}{2})^2 \cdot (0)^3 = (\frac{7}{2})^2 \cdot 0 = 0$$.
Ответ: а) $$2 \frac{11}{26}$$; б) $$2 \frac{3}{4}$$; д) $$10 \frac{2791}{2940}$$; е) 0