Реши примеры с дробями из задания 1а

1) a) $\frac{7}{13} \cdot \frac{26}{49} = \frac{7 \cdot 26}{13 \cdot 49} = \frac{7 \cdot 13 \cdot 2}{13 \cdot 7 \cdot 7} = \frac{2}{7}$ 2) a) $1\frac{2}{5} \cdot 1\frac{1}{3} = \frac{7}{5} \cdot \frac{4}{3} = \frac{7 \cdot 4}{5 \cdot 3} = \frac{28}{15} = 1\frac{13}{15}$ 3) a) $\frac{1}{7} : (-\frac{1}{49}) = \frac{1}{7} \cdot (-49) = -\frac{49}{7} = -7$ 4) a) $\frac{3}{7} \cdot (-7) = -\frac{3 \cdot 7}{7} = -3$ 6) $\frac{5}{7} : (-\frac{3}{21}) = \frac{5}{7} : (-\frac{1}{7}) = \frac{5}{7} \cdot (-7) = -5$ 6) $- \frac{2}{5} : 10 = - \frac{2}{5} \cdot \frac{1}{10} = -\frac{2}{50} = -\frac{1}{25}$ б) $-6\frac{1}{4} \cdot 1\frac{3}{5} = -\frac{25}{4} \cdot \frac{8}{5} = -\frac{25 \cdot 8}{4 \cdot 5} = -\frac{5 \cdot 5 \cdot 4 \cdot 2}{4 \cdot 5} = -10$ в) $-3\frac{1}{5} \cdot (-1\frac{1}{4}) = -\frac{16}{5} \cdot (-\frac{5}{4}) = \frac{16 \cdot 5}{5 \cdot 4} = \frac{4 \cdot 4 \cdot 5}{5 \cdot 4} = 4$ б) $\frac{5}{14} : \frac{42}{75} = \frac{5}{14} \cdot \frac{75}{42} = \frac{5 \cdot 75}{14 \cdot 42} = \frac{5 \cdot 3 \cdot 25}{2 \cdot 7 \cdot 6 \cdot 7} = \frac{5 \cdot 3 \cdot 25}{2 \cdot 7 \cdot 2 \cdot 3 \cdot 7} = \frac{5 \cdot 25}{2 \cdot 7 \cdot 2 \cdot 7} = \frac{125}{196}$ в) $\frac{13}{24} \cdot (-\frac{48}{39}) = -\frac{13 \cdot 48}{24 \cdot 39} = -\frac{13 \cdot 24 \cdot 2}{24 \cdot 13 \cdot 3} = -\frac{2}{3}$ б) $3\frac{5}{7} : 5\frac{4}{7} = \frac{26}{7} : \frac{39}{7} = \frac{26}{7} \cdot \frac{7}{39} = \frac{26 \cdot 7}{7 \cdot 39} = \frac{2 \cdot 13 \cdot 7}{7 \cdot 3 \cdot 13} = \frac{2}{3}$ в) $-7 : (-2\frac{1}{3}) = -7 : (-\frac{7}{3}) = -7 \cdot (-\frac{3}{7}) = \frac{7 \cdot 3}{7} = 3$