Задание 135 Вариант 3 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\left(9\frac{1}{4}-8\frac{2}{3}\right)\cdot 1\frac{5}{7}+\left(4\frac{2}{9}-2\frac{5}{6}\right):1\frac{1}{9}$$
$$9\frac{1}{4}-8\frac{2}{3}=9\frac{3}{12}-8\frac{8}{12}=8\frac{15}{12}-8\frac{8}{12}=\frac{7}{12}$$
$$4\frac{2}{9}-2\frac{5}{6}=4\frac{4}{18}-2\frac{15}{18}=3\frac{22}{18}-2\frac{15}{18}=1\frac{7}{18}$$
$$\frac{7}{12}\cdot 1\frac{5}{7}=\frac{7}{12}\cdot \frac{12}{7}=1$$
$$1\frac{7}{18}:1\frac{1}{9}=\frac{25}{18}:\frac{10}{9}=\frac{25}{18}\cdot \frac{9}{10}=\frac{25\cdot 9}{18\cdot 10}=\frac{5\cdot 1}{2\cdot 2}=\frac{5}{4}=1\frac{1}{4}$$
$$1+1\frac{1}{4}=2\frac{1}{4}$$

б)

$$2\frac{3}{11}\cdot \frac{7}{9}+6\frac{8}{11}\cdot \frac{9}{7}-1\frac{1}{8}$$
$$2\frac{3}{11}\cdot \frac{7}{9}+6\frac{8}{11}\cdot \frac{9}{7}-1\frac{1}{8} = 2\frac{3}{11}\cdot \frac{7}{9}+6\frac{8}{11}\cdot \frac{7}{9}-1\frac{1}{8}$$
$$=\left(2\frac{3}{11}+6\frac{8}{11}\right)\cdot \frac{7}{9}-1\frac{1}{8}$$
$$=8\frac{11}{11}\cdot \frac{7}{9}-1\frac{1}{8} =9\cdot \frac{7}{9}-1\frac{1}{8} =7-1\frac{1}{8} =6\frac{7}{8}$$

Ответ: а) $$2\frac{1}{4}$$; б) $$6\frac{7}{8}$$.