Задание 2 (Вариант I): Вычислите: а) \(\frac{5}{7} \cdot (\frac{21}{20} - \frac{7}{30}) + \frac{16}{21} \cdot \frac{8}{7}\), б) \((1 \frac{1}{2} - \frac{1}{3})^3 : (1 \frac{1}{3} - \frac{1}{4})^2 \cdot (\frac{3}{2})^2\)

а) \(\frac{5}{7} \cdot (\frac{21}{20} - \frac{7}{30}) + \frac{16}{21} \cdot \frac{8}{7} = \frac{5}{7} \cdot (\frac{63 - 14}{60}) + \frac{128}{147} = \frac{5}{7} \cdot \frac{49}{60} + \frac{128}{147} = \frac{245}{420} + \frac{128}{147} = \frac{7}{12} + \frac{128}{147} = \frac{343 + 512}{588} = \frac{855}{588} = \frac{285}{196}\) б) \((1 \frac{1}{2} - \frac{1}{3})^3 : (1 \frac{1}{3} - \frac{1}{4})^2 \cdot (\frac{3}{2})^2 = (\frac{3}{2} - \frac{1}{3})^3 : (\frac{4}{3} - \frac{1}{4})^2 \cdot \frac{9}{4} = (\frac{9 - 2}{6})^3 : (\frac{16 - 3}{12})^2 \cdot \frac{9}{4} = (\frac{7}{6})^3 : (\frac{13}{12})^2 \cdot \frac{9}{4} = \frac{343}{216} : \frac{169}{144} \cdot \frac{9}{4} = \frac{343}{216} \cdot \frac{144}{169} \cdot \frac{9}{4} = \frac{343 \cdot 144 \cdot 9}{216 \cdot 169 \cdot 4} = \frac{343 \cdot 2 \cdot 1}{6 \cdot 169 \cdot 1} = \frac{686}{1014} = \frac{343}{507}\)