Упр.635 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)

Задача

Вычисли:
1) $$\frac{\left(12\frac{4}{5}\cdot3\frac{3}{4}-4\frac{4}{11}\cdot4\frac{1}{8}\right)}{\left(11\frac{2}{3}\cdot2\frac{4}{7}:4\frac{1}{4}\right)}$$;
2) $$\frac{\left(9\frac{4}{5}:4\frac{2}{3}+3\frac{1}{3}:1\frac{1}{3}\right)}{\left(5\frac{3}{4}:2\frac{1}{2}\right)}$$;
3) $$\frac{\left(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}\right)\cdot3\frac{3}{5}:2\frac{1}{2}}{14-15\frac{1}{8}:2\frac{1}{5}}$$.

Подробный ответ

1) Преобразуем смешанные числа в неправильные дроби и выполняем действия по порядку:

$$\frac{12\frac{4}{5}\cdot 3\frac{3}{4}-4\frac{4}{11}\cdot 4\frac{1}{8}}{11\frac{2}{3}\cdot 2\frac{4}{7}:4\frac{1}{4}} = \frac{\frac{64}{5}\cdot \frac{15}{4}-\frac{48}{11}\cdot \frac{33}{8}}{\frac{35}{3}\cdot \frac{18}{7}:\frac{17}{4}}$$

$$\frac{\frac{64\cdot 15}{5\cdot 4}-\frac{48\cdot 33}{11\cdot 8}}{\frac{35\cdot 18}{3\cdot 7}\cdot \frac{4}{17}} = \frac{48-18}{\frac{120}{17}} = 30:\frac{120}{17} = 30\cdot \frac{17}{120} = \frac{17}{4} = 4\frac{1}{4}$$

2) Аналогично:

$$\frac{9\frac{4}{5}:4\frac{2}{3}+3\frac{1}{3}:1\frac{1}{3}}{5\frac{3}{4}:2\frac{1}{2}} = \frac{\frac{49}{5}:\frac{14}{3}+\frac{10}{3}:\frac{4}{3}}{\frac{23}{4}:\frac{5}{2}}$$

$$\frac{\frac{49}{5}\cdot \frac{3}{14}+\frac{10}{3}\cdot \frac{3}{4}}{\frac{23}{4}\cdot \frac{2}{5}} = \frac{\frac{21}{10}+\frac{5}{2}}{\frac{23}{10}} = \frac{\frac{21}{10}+\frac{25}{10}}{\frac{23}{10}} = \frac{\frac{46}{10}}{\frac{23}{10}} = \frac{46}{10}:\frac{23}{10} = 2$$

3) Сначала вычислим сумму в числителе и произведение в знаменателе:

$$\frac{\left(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}\right)\cdot 3\frac{3}{5}:2\frac{1}{2}}{14-15\frac{1}{8}:2\frac{1}{5}}$$

$$1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4} = \frac{3}{2}+\frac{8}{3}+\frac{15}{4} = \frac{18+32+45}{12} = \frac{95}{12}$$

$$3\frac{3}{5}:2\frac{1}{2} = \frac{18}{5}:\frac{5}{2} = \frac{18}{5}\cdot \frac{2}{5} = \frac{36}{25}$$

$$15\frac{1}{8}:2\frac{1}{5} = \frac{121}{8}:\frac{11}{5} = \frac{121}{8}\cdot \frac{5}{11} = \frac{55}{8}$$

Тогда
$$\frac{\frac{95}{12}\cdot \frac{36}{25}}{14-\frac{55}{8}} = \frac{\frac{57}{5}}{7\frac{1}{8}} = \frac{\frac{57}{5}}{\frac{57}{8}} = \frac{57}{5}:\frac{57}{8} = \frac{8}{5} = 1\frac{3}{5}$$