Задание 109 Вариант 3 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\left(4\frac{1}{12}\cdot 1\frac{5}{7}-5\frac{2}{9}\right)\cdot 3\frac{3}{8}$$
$$4\frac{1}{12}=\frac{49}{12},\qquad 1\frac{5}{7}=\frac{12}{7}$$
$$\frac{49}{12}\cdot \frac{12}{7}=7$$
$$5\frac{2}{9}=5+\frac{2}{9}=6\frac{9}{9}-\frac{2}{9}=6\frac{7}{9}$$
$$7-5\frac{2}{9}=1\frac{7}{9}=\frac{16}{9}$$
$$3\frac{3}{8}=\frac{27}{8}$$
$$\frac{16}{9}\cdot \frac{27}{8}=\frac{2}{3}\cdot 3=2$$

б)

$$\left(\frac{3}{8}+2\frac{2}{7}\cdot 1\frac{1}{20}\right)\cdot 3\frac{1}{3}$$
$$2\frac{2}{7}=\frac{16}{7},\qquad 1\frac{1}{20}=\frac{21}{20}$$
$$\frac{16}{7}\cdot \frac{21}{20}=\frac{4\cdot 3}{5}=\frac{12}{5}=2\frac{2}{5}$$
$$\frac{3}{8}+2\frac{2}{5}=\frac{3}{8}+\frac{12}{5}=\frac{15}{40}+\frac{96}{40}=\frac{111}{40}=2\frac{31}{40}$$
$$3\frac{1}{3}=\frac{10}{3}$$
$$\frac{111}{40}\cdot \frac{10}{3}=\frac{111}{12}=\frac{37}{4}=9\frac{1}{4}$$

Ответ: а) $$2$$; б) $$9\frac{1}{4}$$.