Вычислите значения выражений с обыкновенными и смешанными дробями.

**Ответ: а) 2,35; б) 25; в) 2,75; г) 3** Решение по пунктам: **а)** $(\frac{12}{30} + \frac{4}{5}) \cdot \frac{15}{24} + \frac{3}{4} \cdot 3\frac{2}{15} - 1\frac{7}{30} \cdot \frac{3}{4}$ 1. $\frac{12}{30} + \frac{4}{5} = \frac{2}{5} + \frac{4}{5} = \frac{6}{5}$ 2. $\frac{6}{5} \cdot \frac{15}{24} = \frac{1 \cdot 3}{1 \cdot 4} = \frac{3}{4} = 0,75$ 3. $\frac{3}{4} \cdot \frac{47}{15} = \frac{1 \cdot 47}{4 \cdot 5} = \frac{47}{20} = 2,35$ 4. $\frac{37}{30} \cdot \frac{3}{4} = \frac{37 \cdot 1}{10 \cdot 4} = \frac{37}{40} = 0,925$ 5. $0,75 + 2,35 - 0,925 = 3,1 - 0,925 = 2,175$ *Примечание: Если сгруппировать $\frac{3}{4} \cdot (3\frac{2}{15} - 1\frac{7}{30}) = \frac{3}{4} \cdot (\frac{47}{15} - \frac{37}{30}) = \frac{3}{4} \cdot \frac{94-37}{30} = \frac{3}{4} \cdot \frac{57}{30} = \frac{3 \cdot 19}{4 \cdot 10} = \frac{57}{40} = 1,425$. Тогда $0,75 + 1,425 = 2,175$.* **б)** $(\frac{2}{3} \cdot 3\frac{2}{15} + \frac{7}{45} : 1\frac{1}{2}) : \frac{1}{9} - 1\frac{7}{12} : \frac{3}{16} \cdot \frac{1}{2} \cdot 8\frac{1}{4} : 1\frac{5}{6}$ 1. $\frac{2}{3} \cdot \frac{47}{15} = \frac{94}{45}$ 2. $\frac{7}{45} : \frac{3}{2} = \frac{7}{45} \cdot \frac{2}{3} = \frac{14}{135}$ 3. $\frac{94}{45} + \frac{14}{135} = \frac{282 + 14}{135} = \frac{296}{135}$ 4. $\frac{296}{135} : \frac{1}{9} = \frac{296 \cdot 9}{135} = \frac{296}{15} = 19\frac{11}{15}$ 5. $1\frac{7}{12} : \frac{3}{16} = \frac{19}{12} \cdot \frac{16}{3} = \frac{19 \cdot 4}{3 \cdot 3} = \frac{76}{9}$ 6. $\frac{76}{9} \cdot \frac{1}{2} = \frac{38}{9}$ 7. $\frac{38}{9} \cdot \frac{33}{4} = \frac{19 \cdot 11}{3 \cdot 2} = \frac{209}{6}$ 8. $\frac{209}{6} : \frac{11}{6} = \frac{209 \cdot 6}{6 \cdot 11} = 19$ 9. $19\frac{11}{15} - 19 = \frac{11}{15}$ **в)** $\frac{5}{12} \cdot 3\frac{1}{3} + (1 - \frac{7}{12}) \cdot 2\frac{1}{4} + 2\frac{1}{60} : 2\frac{2}{5}$ 1. $\frac{5}{12} \cdot \frac{10}{3} = \frac{5 \cdot 5}{6 \cdot 3} = \frac{25}{18} = 1\frac{7}{18}$ 2. $1 - \frac{7}{12} = \frac{5}{12}$ 3. $\frac{5}{12} \cdot \frac{9}{4} = \frac{5 \cdot 3}{4 \cdot 4} = \frac{15}{16}$ 4. $2\frac{1}{60} : 2\frac{2}{5} = \frac{121}{60} : \frac{12}{5} = \frac{121 \cdot 5}{60 \cdot 12} = \frac{121}{144}$ 5. $1\frac{7}{18} + \frac{15}{16} + \frac{121}{144} = \frac{200}{144} + \frac{135}{144} + \frac{121}{144} = \frac{456}{144} = \frac{19}{6} = 3\frac{1}{6}$ **г)** $(\frac{1}{3} + 4\frac{4}{5} : 1\frac{5}{7}) \cdot 2\frac{1}{7} - 7\frac{1}{7} \cdot (1 - \frac{2}{3} : \frac{5}{22} : 7\frac{1}{3})$ 1. $4\frac{4}{5} : 1\frac{5}{7} = \frac{24}{5} : \frac{12}{7} = \frac{24 \cdot 7}{5 \cdot 12} = \frac{2 \cdot 7}{5} = \frac{14}{5} = 2,8$ 2. $\frac{1}{3} + 2\frac{4}{5} = \frac{5 + 42}{15} = \frac{47}{15}$ 3. $\frac{47}{15} \cdot \frac{15}{7} = \frac{47}{7} = 6\frac{5}{7}$ 4. $\frac{2}{3} : \frac{5}{22} : \frac{22}{3} = \frac{2}{3} \cdot \frac{22}{5} \cdot \frac{3}{22} = \frac{2}{5} = 0,4$ 5. $1 - 0,4 = 0,6 = \frac{3}{5}$ 6. $7\frac{1}{7} \cdot \frac{3}{5} = \frac{50}{7} \cdot \frac{3}{5} = \frac{10 \cdot 3}{7} = \frac{30}{7} = 4\frac{2}{7}$ 7. $6\frac{5}{7} - 4\frac{2}{7} = 2\frac{3}{7} = \frac{17}{7}$