Упр.80 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 6 класс (Математика)

Найдите значение выражения:

1) $$4\frac{1}{5}:\frac{7}{15}:\frac{1}{3}$$;

3) $$\left(5\frac{8}{9}:1\frac{17}{36}+1\frac{1}{4}\right):4\frac{1}{5}$$;

2) $$4\frac{1}{5}:\left(\frac{7}{15}:\frac{1}{3}\right)$$;

4) $$\left(2\frac{1}{4}+4\frac{5}{6}\right):3\frac{2}{5}-\frac{3}{4}:\frac{3}{5}$$.

1) $$4\frac{1}{5} : \frac{7}{15} : \frac{1}{3} = \frac{21}{5} \cdot \frac{15}{7} \cdot 3 = \frac{21 \cdot 15 \cdot 3}{5 \cdot 7} = 27.$$

2) $$4\frac{1}{5} : \left(\frac{7}{15} : \frac{1}{3}\right) = \frac{21}{5} : \left(\frac{7}{15} \cdot 3\right) = \frac{21}{5} : \frac{7}{5} = \frac{21}{5} \cdot \frac{5}{7} = 3.$$

3) $$\left(5\frac{8}{9} : 1\frac{17}{36} + 1\frac{1}{4}\right) : 4\frac{1}{5} = \left(\frac{53}{9} : \frac{53}{36} + 1\frac{1}{4}\right) : \frac{21}{5}$$

$$= \left(\frac{53}{9} \cdot \frac{36}{53} + 1\frac{1}{4}\right) : \frac{21}{5} = \left(4 + 1\frac{1}{4}\right) : \frac{21}{5} = 5\frac{1}{4} : \frac{21}{5}$$

$$= \frac{21}{4} \cdot \frac{5}{21} = \frac{5}{4} = 1\frac{1}{4}.$$

4) $$\left(2\frac{1}{4} + 4\frac{5}{6}\right) : 3\frac{2}{5} — \frac{3}{4} : \frac{3}{5} = \left(2\frac{3}{12} + 4\frac{10}{12}\right) : \frac{17}{5} — \frac{3}{4} \cdot \frac{5}{3}$$

$$= 6\frac{13}{12} : \frac{17}{5} — \frac{5}{4} = \frac{85}{12} \cdot \frac{5}{17} — \frac{5}{4} = \frac{25}{12} — \frac{5}{4}$$

$$= \frac{25}{12} — \frac{15}{12} = \frac{10}{12} = \frac{5}{6}.$$

Ответ: 1) 27; 2) 3; 3) $$1\frac{1}{4}$$; 4) $$\frac{5}{6}$$.