Задание 252 Вариант 1 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$-15{,}64 : 4\frac{3}{5} + 7{,}1 \cdot 2\frac{2}{5}$$
$$-15{,}64 : 4\frac{3}{5} = -15{,}64 : 4{,}6 = -3{,}4$$
$$-3{,}4 + 7{,}1 = 3{,}7$$
$$3{,}7 \cdot 2\frac{2}{5} = 3{,}7 \cdot \frac{12}{5} = \frac{37}{10}\cdot\frac{12}{5}=\frac{222}{25}=8{,}88$$

$$-15{,}64 : 4\frac{3}{5} + 7{,}1 \cdot 2\frac{2}{5} = 8{,}88$$

б)

$$\left(0{,}4-\frac{11}{15}\right)\cdot 3\frac{2}{7}-\left(\frac{7}{18}-0{,}5\right):4\frac{1}{6}$$
$$0{,}4-\frac{11}{15}=\frac{2}{5}-\frac{11}{15}=\frac{6}{15}-\frac{11}{15}=-\frac{5}{15}=-\frac{1}{3}$$
$$\frac{7}{18}-0{,}5=\frac{7}{18}-\frac{1}{2}=\frac{7}{18}-\frac{9}{18}=-\frac{2}{18}=-\frac{1}{9}$$
$$-\frac{1}{3}\cdot 3\frac{2}{7}=-\frac{1}{3}\cdot\frac{23}{7}=-\frac{23}{21}$$
$$-\frac{1}{9}:4\frac{1}{6}=-\frac{1}{9}:\frac{25}{6}=-\frac{1}{9}\cdot\frac{6}{25}=-\frac{2}{75}$$
$$-\frac{23}{21}-\left(-\frac{2}{75}\right)=-\frac{23}{21}+\frac{2}{75}=-\frac{575}{525}+\frac{14}{525}=-\frac{561}{525}=-\frac{187}{175}$$

$$\left(0{,}4-\frac{11}{15}\right)\cdot 3\frac{2}{7}-\left(\frac{7}{18}-0{,}5\right):4\frac{1}{6}=-\frac{187}{175}$$