Решай три дня в неделю по 1 – 2 различных заданий. Оформляй задания правильно: столбики с вычислениями записывай в тетради, к задачам делай схему или краткую запись условия.

Давай разберем эти примеры на действия с обыкновенными дробями. Помни главное правило: сначала выполняем действия в скобках, затем умножение и деление, и в конце — сложение и вычитание. ### Решение: 1) $14 - (10 \frac{11}{13} - 5 \frac{9}{13}) \cdot 2 = 14 - (5 \frac{2}{13}) \cdot 2 = 14 - \frac{67}{13} \cdot 2 = 14 - \frac{134}{13} = 14 - 10 \frac{4}{13} = 3 \frac{9}{13}$ 2) $24 \frac{8}{41} : 4 - 18 \frac{6}{41} : 3 = \frac{992}{41} \cdot \frac{1}{4} - \frac{744}{41} \cdot \frac{1}{3} = \frac{248}{41} - \frac{248}{41} = 0$ 3) $\frac{27 \frac{3}{8} - 21 \frac{7}{20}}{(3 \frac{4}{7} - 1 \frac{23}{28}) - (1 \frac{47}{65} - \frac{29}{130})} = \frac{\frac{219}{8} - \frac{427}{20}}{(\frac{25}{7} - \frac{51}{28}) - (\frac{112}{65} - \frac{29}{130})} = \frac{\frac{1095-854}{40}}{(\frac{100-51}{28}) - (\frac{224-29}{130})} = \frac{\frac{241}{40}}{\frac{49}{28} - \frac{195}{130}} = \frac{\frac{241}{40}}{\frac{7}{4} - \frac{3}{2}} = \frac{\frac{241}{40}}{\frac{7-6}{4}} = \frac{241}{40} : \frac{1}{4} = \frac{241}{10} = 24,1$ 4) $[(1 \frac{1}{2} + 2 \frac{2}{3}) : 3 \frac{3}{4} - 2 \frac{4}{5}] \cdot 1 \frac{8}{9} + \frac{1}{4} = [(\frac{3}{2} + \frac{8}{3}) : \frac{15}{4} - \frac{14}{5}] \cdot \frac{17}{9} + \frac{1}{4} = [(\frac{9+16}{6}) : \frac{15}{4} - \frac{14}{5}] \cdot \frac{17}{9} + \frac{1}{4} = [\frac{25}{6} \cdot \frac{4}{15} - \frac{14}{5}] \cdot \frac{17}{9} + \frac{1}{4} = [\frac{10}{9} - \frac{14}{5}] \cdot \frac{17}{9} + \frac{1}{4} = [\frac{50-126}{45}] \cdot \frac{17}{9} + \frac{1}{4} = -\frac{76}{45} \cdot \frac{17}{9} + \frac{1}{4} = -\frac{1292}{405} + \frac{1}{4} = -3 \frac{77}{405} + 0,25 = -\frac{1292}{405} + \frac{1}{4} = \frac{-5168 + 405}{1620} = -\frac{4763}{1620} \approx -2,94$ 5) $(3 \frac{1}{4} + 2 \frac{1}{6}) : 2 \frac{3}{5} - 2 \cdot 2 \frac{1}{4} + 5 \frac{1}{6} = (\frac{13}{4} + \frac{13}{6}) : \frac{13}{5} - 2 \cdot \frac{9}{4} + \frac{31}{6} = (\frac{39+26}{12}) \cdot \frac{5}{13} - \frac{9}{2} + \frac{31}{6} = \frac{65}{12} \cdot \frac{5}{13} - \frac{9}{2} + \frac{31}{6} = \frac{25}{12} - \frac{54}{12} + \frac{62}{12} = \frac{33}{12} = \frac{11}{4} = 2,75$ 6) $[1 \frac{1}{10} + 7 : (3 \frac{1}{2} - 1 \frac{5}{12})] \cdot 1 \frac{1}{59} = [\frac{11}{10} + 7 : (\frac{7}{2} - \frac{17}{12})] \cdot \frac{60}{59} = [\frac{11}{10} + 7 : (\frac{42-17}{12})] \cdot \frac{60}{59} = [\frac{11}{10} + 7 : \frac{25}{12}] \cdot \frac{60}{59} = [\frac{11}{10} + 7 \cdot \frac{12}{25}] \cdot \frac{60}{59} = [\frac{11}{10} + \frac{84}{25}] \cdot \frac{60}{59} = [\frac{55+168}{50}] \cdot \frac{60}{59} = \frac{223}{50} \cdot \frac{60}{59} = \frac{223 \cdot 6}{5 \cdot 59} = \frac{1338}{295} \approx 4,53$ 7) $(7 \frac{1}{3} - 6 \frac{7}{8}) : 3 \frac{3}{4} - (5 \frac{1}{4} - 4 \frac{21}{40}) : 1 \frac{9}{20} = (\frac{22}{3} - \frac{55}{8}) : \frac{15}{4} - (\frac{21}{4} - \frac{181}{40}) : \frac{29}{20} = (\frac{176-165}{24}) : \frac{15}{4} - (\frac{210-181}{40}) : \frac{29}{20} = \frac{11}{24} \cdot \frac{4}{15} - \frac{29}{40} \cdot \frac{20}{29} = \frac{11}{6 \cdot 15} - \frac{1}{2} = \frac{11}{90} - \frac{45}{90} = -\frac{34}{90} = -\frac{17}{45}$ 8) $(\frac{2}{15} + 1 \frac{7}{12}) \cdot \frac{30}{103} - 2 : 2 \frac{1}{4} \cdot \frac{9}{32} = (\frac{2}{15} + \frac{19}{12}) \cdot \frac{30}{103} - 2 : \frac{9}{4} \cdot \frac{9}{32} = (\frac{8+95}{60}) \cdot \frac{30}{103} - 2 \cdot \frac{4}{9} \cdot \frac{9}{32} = \frac{103}{60} \cdot \frac{30}{103} - \frac{8}{9} \cdot \frac{9}{32} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} = 0,25$ 9) $3 \frac{1}{8} : [(4 \frac{5}{12} - 2 \frac{13}{24}) \cdot \frac{4}{7} + (3 \frac{1}{18} - 2 \frac{7}{12}) \cdot 1 \frac{10}{17}] = \frac{25}{8} : [(\frac{53}{12} - \frac{61}{24}) \cdot \frac{4}{7} + (\frac{55}{18} - \frac{31}{12}) \cdot \frac{27}{17}] = \frac{25}{8} : [(\frac{106-61}{24}) \cdot \frac{4}{7} + (\frac{110-93}{36}) \cdot \frac{27}{17}] = \frac{25}{8} : [\frac{45}{24} \cdot \frac{4}{7} + \frac{17}{36} \cdot \frac{27}{17}] = \frac{25}{8} : [\frac{15}{8} \cdot \frac{4}{7} + \frac{3}{4}] = \frac{25}{8} : [\frac{15}{14} + \frac{3}{4}] = \frac{25}{8} : [\frac{30+21}{28}] = \frac{25}{8} : \frac{51}{28} = \frac{25}{8} \cdot \frac{28}{51} = \frac{25 \cdot 7}{2 \cdot 51} = \frac{175}{102} = 1 \frac{73}{102}$ 10) $2 \frac{3}{4} : [(4 \frac{5}{7} - 1 \frac{11}{14}) \cdot 4 \frac{2}{3} + (3 \frac{2}{9} - 1 \frac{5}{6}) \cdot \frac{18}{25}] = \frac{11}{4} : [(\frac{33}{7} - \frac{25}{14}) \cdot \frac{14}{3} + (\frac{29}{9} - \frac{11}{6}) \cdot \frac{18}{25}] = \frac{11}{4} : [(\frac{66-25}{14}) \cdot \frac{14}{3} + (\frac{58-33}{18}) \cdot \frac{18}{25}] = \frac{11}{4} : [\frac{41}{14} \cdot \frac{14}{3} + \frac{25}{18} \cdot \frac{18}{25}] = \frac{11}{4} : [\frac{41}{3} + 1] = \frac{11}{4} : \frac{44}{3} = \frac{11}{4} \cdot \frac{3}{44} = \frac{3}{16} = 0,1875$ Десятичные дроби: 1) $(12 - 8,4) : 0,09 \cdot 0,7 - 0,3 \cdot (0,6 + 3,12) : (14,18 - 7,98) : 0,01 = 3,6 : 0,09 \cdot 0,7 - 0,3 \cdot 3,72 : 6,2 : 0,01 = 40 \cdot 0,7 - 1,116 : 6,2 : 0,01 = 28 - 0,18 : 0,01 = 28 - 18 = 10$