Задание 135 Вариант 2 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\left(6\frac{1}{7}-5\frac{3}{4}\right):\frac{3}{14}+\left(3\frac{3}{4}-1\frac{5}{6}\right):\frac{4}{6}$$
$$6\frac{1}{7}-5\frac{3}{4}=6\frac{4}{28}-5\frac{21}{28}=5\frac{32}{28}-5\frac{21}{28}=\frac{11}{28}$$
$$3\frac{3}{4}-1\frac{5}{6}=3\frac{9}{12}-1\frac{10}{12}=2\frac{21}{12}-1\frac{10}{12}=1\frac{11}{12}$$
$$\frac{11}{28}:\frac{11}{14}=\frac{11}{28}\cdot\frac{14}{11}=\frac{1}{2}$$
$$1\frac{11}{12}:\frac{1}{6}=\frac{23}{12}\cdot\frac{6}{1}=\frac{23}{2}=11\frac{1}{2}$$
$$\frac{1}{2}+11\frac{1}{2}=12$$

б)

$$5\frac{4}{19}\cdot 3\frac{4}{19}+1\frac{15}{19}:\frac{7}{25}-1\frac{2}{3}$$
$$5\frac{4}{19}\cdot 3\frac{4}{19}=\frac{99}{19}\cdot\frac{25}{7}$$
$$1\frac{15}{19}:\frac{7}{25}=\frac{34}{19}\cdot\frac{25}{7}$$
Тогда
$$\frac{99}{19}\cdot\frac{25}{7}+\frac{34}{19}\cdot\frac{25}{7}-1\frac{2}{3} =\frac{25}{7}\left(\frac{99}{19}+\frac{34}{19}\right)-1\frac{2}{3}$$
$$=\frac{25}{7}\cdot\frac{133}{19}-1\frac{2}{3} =\frac{25}{7}\cdot 7-1\frac{2}{3} =25-1\frac{2}{3} =23\frac{1}{3}$$

Ответ: а) $$12$$; б) $$23\frac{1}{3}$$.