a) 1 \frac{1}{3} \cdot 2 = (1 + \frac{1}{3}) \cdot 2 = 1 \cdot 2 + \frac{1}{3} \cdot 2 = 2 + \frac{2}{3} = 2 \frac{2}{3}; б) 1 \frac{1}{5} \cdot 2; в) 2 \frac{1}{5} \cdot 3; г) 3 \frac{1}{4} \cdot 3; д) 2 \frac{2}{7} \cdot 3; e) 2 \cdot 5 \frac{1}{4}; ж) 2 \frac{4}{9} \cdot 9; з) 2 \cdot 5 \frac{7}{8}; и) 2 \frac{1}{9} \cdot 3.

Давай решим эти примеры, используя распределительный закон. б) \(1 \frac{1}{5} \cdot 2\) = \((1 + \frac{1}{5}) \cdot 2\) = \(1 \cdot 2 + \frac{1}{5} \cdot 2\) = \(2 + \frac{2}{5}\) = \(2 \frac{2}{5}\) в) \(2 \frac{1}{5} \cdot 3\) = \((2 + \frac{1}{5}) \cdot 3\) = \(2 \cdot 3 + \frac{1}{5} \cdot 3\) = \(6 + \frac{3}{5}\) = \(6 \frac{3}{5}\) г) \(3 \frac{1}{4} \cdot 3\) = \((3 + \frac{1}{4}) \cdot 3\) = \(3 \cdot 3 + \frac{1}{4} \cdot 3\) = \(9 + \frac{3}{4}\) = \(9 \frac{3}{4}\) д) \(2 \frac{2}{7} \cdot 3\) = \((2 + \frac{2}{7}) \cdot 3\) = \(2 \cdot 3 + \frac{2}{7} \cdot 3\) = \(6 + \frac{6}{7}\) = \(6 \frac{6}{7}\) e) \(2 \cdot 5 \frac{1}{4}\) = \(2 \cdot (5 + \frac{1}{4})\) = \(2 \cdot 5 + 2 \cdot \frac{1}{4}\) = \(10 + \frac{2}{4}\) = \(10 + \frac{1}{2}\) = \(10 \frac{1}{2}\) ж) \(2 \frac{4}{9} \cdot 9\) = \((2 + \frac{4}{9}) \cdot 9\) = \(2 \cdot 9 + \frac{4}{9} \cdot 9\) = \(18 + 4\) = \(22\) з) \(2 \cdot 5 \frac{7}{8}\) = \(2 \cdot (5 + \frac{7}{8})\) = \(2 \cdot 5 + 2 \cdot \frac{7}{8}\) = \(10 + \frac{14}{8}\) = \(10 + \frac{7}{4}\) = \(10 + 1 \frac{3}{4}\) = \(11 \frac{3}{4}\) и) \(2 \frac{1}{9} \cdot 3\) = \((2 + \frac{1}{9}) \cdot 3\) = \(2 \cdot 3 + \frac{1}{9} \cdot 3\) = \(6 + \frac{3}{9}\) = \(6 + \frac{1}{3}\) = \(6 \frac{1}{3}\)

Ответ: б) \(2 \frac{2}{5}\); в) \(6 \frac{3}{5}\); г) \(9 \frac{3}{4}\); д) \(6 \frac{6}{7}\); e) \(10 \frac{1}{2}\); ж) \(22\); з) \(11 \frac{3}{4}\); и) \(6 \frac{1}{3}\)