Реши пример: 9/56 - (7/15 - 5/12) * (3/14 + 1/2)

Решаю по порядку все примеры. Ты можешь выбрать только те, которые тебе нужны. а) $\frac{9}{56} - (\frac{7}{15} - \frac{5}{12}) \cdot (\frac{3}{14} + \frac{1}{2})$ 1. $\frac{7}{15} - \frac{5}{12} = \frac{7 \cdot 4}{15 \cdot 4} - \frac{5 \cdot 5}{12 \cdot 5} = \frac{28}{60} - \frac{25}{60} = \frac{3}{60} = \frac{1}{20}$ 2. $\frac{3}{14} + \frac{1}{2} = \frac{3}{14} + \frac{1 \cdot 7}{2 \cdot 7} = \frac{3}{14} + \frac{7}{14} = \frac{10}{14} = \frac{5}{7}$ 3. $\frac{1}{20} \cdot \frac{5}{7} = \frac{1 \cdot 5}{20 \cdot 7} = \frac{5}{140} = \frac{1}{28}$ 4. $\frac{9}{56} - \frac{1}{28} = \frac{9}{56} - \frac{1 \cdot 2}{28 \cdot 2} = \frac{9}{56} - \frac{2}{56} = \frac{7}{56} = \frac{1}{8}$ **Ответ: $\frac{1}{8}$** в) $(2 \frac{1}{2})^2 \cdot \frac{8}{15} - \frac{5}{9}$ 1. $2 \frac{1}{2} = \frac{2 \cdot 2 + 1}{2} = \frac{5}{2}$ 2. $(\frac{5}{2})^2 = \frac{5^2}{2^2} = \frac{25}{4}$ 3. $\frac{25}{4} \cdot \frac{8}{15} = \frac{25 \cdot 8}{4 \cdot 15} = \frac{5 \cdot 2}{1 \cdot 3} = \frac{10}{3}$ 4. $\frac{10}{3} - \frac{5}{9} = \frac{10 \cdot 3}{3 \cdot 3} - \frac{5}{9} = \frac{30}{9} - \frac{5}{9} = \frac{25}{9} = 2 \frac{7}{9}$ **Ответ: $2 \frac{7}{9}$** г) $(2 \frac{1}{3} \cdot 1 \frac{2}{7})^3 \cdot \frac{2}{9}$ 1. $2 \frac{1}{3} = \frac{2 \cdot 3 + 1}{3} = \frac{7}{3}$ 2. $1 \frac{2}{7} = \frac{1 \cdot 7 + 2}{7} = \frac{9}{7}$ 3. $\frac{7}{3} \cdot \frac{9}{7} = \frac{7 \cdot 9}{3 \cdot 7} = \frac{1 \cdot 3}{1 \cdot 1} = \frac{3}{1} = 3$ 4. $3^3 = 3 \cdot 3 \cdot 3 = 27$ 5. $27 \cdot \frac{2}{9} = \frac{27 \cdot 2}{9} = \frac{3 \cdot 2}{1} = 6$ **Ответ: 6** д) $( \frac{2}{3} + \frac{7}{8} - \frac{5}{6}) \cdot (1 - \frac{5}{17})$ 1. $\frac{2}{3} + \frac{7}{8} - \frac{5}{6} = \frac{2 \cdot 8}{3 \cdot 8} + \frac{7 \cdot 3}{8 \cdot 3} - \frac{5 \cdot 4}{6 \cdot 4} = \frac{16}{24} + \frac{21}{24} - \frac{20}{24} = \frac{16 + 21 - 20}{24} = \frac{17}{24}$ 2. $1 - \frac{5}{17} = \frac{17}{17} - \frac{5}{17} = \frac{12}{17}$ 3. $\frac{17}{24} \cdot \frac{12}{17} = \frac{17 \cdot 12}{24 \cdot 17} = \frac{1 \cdot 1}{2 \cdot 1} = \frac{1}{2}$ **Ответ: $\frac{1}{2}$** e) $(3 \frac{1}{14} - 2 \frac{5}{21}) \cdot (2,7 - 2,1)$ 1. $3 \frac{1}{14} = \frac{3 \cdot 14 + 1}{14} = \frac{43}{14}$ 2. $2 \frac{5}{21} = \frac{2 \cdot 21 + 5}{21} = \frac{47}{21}$ 3. $\frac{43}{14} - \frac{47}{21} = \frac{43 \cdot 3}{14 \cdot 3} - \frac{47 \cdot 2}{21 \cdot 2} = \frac{129}{42} - \frac{94}{42} = \frac{35}{42} = \frac{5}{6}$ 4. $2,7 - 2,1 = 0,6$ 5. $\frac{5}{6} \cdot 0,6 = \frac{5}{6} \cdot \frac{6}{10} = \frac{5 \cdot 6}{6 \cdot 10} = \frac{1 \cdot 1}{1 \cdot 2} = \frac{1}{2}$ **Ответ: $\frac{1}{2}$** ж) $\frac{7}{11} \cdot ((\frac{3}{7})^2 + \frac{5}{7})$ 1. $(\frac{3}{7})^2 = \frac{3^2}{7^2} = \frac{9}{49}$ 2. $\frac{9}{49} + \frac{5}{7} = \frac{9}{49} + \frac{5 \cdot 7}{7 \cdot 7} = \frac{9}{49} + \frac{35}{49} = \frac{44}{49}$ 3. $\frac{7}{11} \cdot \frac{44}{49} = \frac{7 \cdot 44}{11 \cdot 49} = \frac{1 \cdot 4}{1 \cdot 7} = \frac{4}{7}$ **Ответ: $\frac{4}{7}$** з) $(4 \frac{13}{18} - 3 \frac{7}{9}) \cdot (\frac{1}{2} - \frac{4}{17})$ 1. $4 \frac{13}{18} = \frac{4 \cdot 18 + 13}{18} = \frac{85}{18}$ 2. $3 \frac{7}{9} = \frac{3 \cdot 9 + 7}{9} = \frac{34}{9}$ 3. $\frac{85}{18} - \frac{34}{9} = \frac{85}{18} - \frac{34 \cdot 2}{9 \cdot 2} = \frac{85}{18} - \frac{68}{18} = \frac{17}{18}$ 4. $\frac{1}{2} - \frac{4}{17} = \frac{1 \cdot 17}{2 \cdot 17} - \frac{4 \cdot 2}{17 \cdot 2} = \frac{17}{34} - \frac{8}{34} = \frac{9}{34}$ 5. $\frac{17}{18} \cdot \frac{9}{34} = \frac{17 \cdot 9}{18 \cdot 34} = \frac{1 \cdot 1}{2 \cdot 2} = \frac{1}{4}$ **Ответ: $\frac{1}{4}$** и) $(\frac{7}{12} - \frac{3}{16} - \frac{5}{24}) + (\frac{1}{4})^2$ 1. $\frac{7}{12} - \frac{3}{16} - \frac{5}{24} = \frac{7 \cdot 4}{12 \cdot 4} - \frac{3 \cdot 3}{16 \cdot 3} - \frac{5 \cdot 2}{24 \cdot 2} = \frac{28}{48} - \frac{9}{48} - \frac{10}{48} = \frac{28 - 9 - 10}{48} = \frac{9}{48} = \frac{3}{16}$ 2. $(\frac{1}{4})^2 = \frac{1^2}{4^2} = \frac{1}{16}$ 3. $\frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$ **Ответ: $\frac{1}{4}$**