Упр.80 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 6 класс (Математика)

Найдите значение выражения:

1. 1) $$2\frac{2}{3}: \frac{2}{9}: \frac{1}{4}$$;
2. 3) $$\left(7\frac{5}{7}:3\frac{3}{5}-\frac{1}{7}\right):1\frac{1}{3}$$;
3. 2) $$2\frac{2}{3}:\left(\frac{2}{9}:\frac{1}{4}\right)$$;
4. 4) $$\left(4\frac{5}{12}-3\frac{13}{24}\right):1\frac{3}{4}+\frac{5}{6}:\frac{5}{7}$$.

1) $$2\frac{2}{3} : \frac{2}{9} : \frac{1}{4} = \frac{8}{3} : \frac{2}{9} : \frac{1}{4} = \frac{8}{3}\cdot\frac{9}{2}\cdot 4 = 48$$

2) $$2\frac{2}{3} : \left(\frac{2}{9} : \frac{1}{4}\right) = \frac{8}{3} : \left(\frac{2}{9}\cdot 4\right) = \frac{8}{3} : \frac{8}{9} = \frac{8}{3}\cdot\frac{9}{8} = 3$$

3) $$\left(7\frac{5}{7} : 3\frac{3}{5} — \frac{1}{7}\right) : 1\frac{1}{3} = \left(\frac{54}{7} : \frac{18}{5} — \frac{1}{7}\right) : \frac{4}{3}$$

$$= \left(\frac{54}{7}\cdot\frac{5}{18} — \frac{1}{7}\right)\cdot\frac{3}{4} = \left(\frac{15}{7} — \frac{1}{7}\right)\cdot\frac{3}{4} = \frac{14}{7}\cdot\frac{3}{4} = 2\cdot\frac{3}{4} = \frac{3}{2} = 1{,}5$$

4) $$\left(4\frac{5}{12} — 3\frac{13}{24}\right) : 1\frac{3}{4} + \frac{5}{6} : \frac{5}{7} = \left(4\frac{10}{24} — 3\frac{13}{24}\right) : \frac{7}{4} + \frac{5}{6}\cdot\frac{7}{5}$$

$$= \left(3\frac{34}{24} — 3\frac{13}{24}\right)\cdot\frac{4}{7} + \frac{7}{6} = \frac{21}{24}\cdot\frac{4}{7} + \frac{7}{6} = \frac{7}{8}\cdot\frac{4}{7} + \frac{7}{6} = \frac{1}{2} + \frac{7}{6}$$

$$= \frac{3}{6} + \frac{7}{6} = \frac{10}{6} = \frac{5}{3} = 1\frac{2}{3}$$

Ответ: 1) 48; 2) 3; 3) 1,5; 4) $$1\frac{2}{3}$$.