Задание 252 Вариант 2 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\left(-48\frac{3}{4}:3{,}9+8\frac{1}{5}\right)\cdot(-6{,}3)$$
$$-48\frac{3}{4}:3{,}9=-\frac{195}{4}:\frac{39}{10}=-\frac{195}{4}\cdot\frac{10}{39}=-\frac{25}{2}=-12\frac{1}{2}$$
$$-12\frac{1}{2}+8\frac{1}{5}=-12\frac{5}{10}+8\frac{2}{10}=-4\frac{3}{10}=-4{,}3$$
$$-4{,}3\cdot(-6{,}3)=27{,}09$$

б)

$$\left(0{,}2-1\frac{7}{15}\right)\cdot\left(-\frac{5}{8}\right)-5\left(-\frac{5}{12}-2{,}75\right):4\frac{1}{3}$$
$$0{,}2-1\frac{7}{15}=\frac{1}{5}-\frac{22}{15}=\frac{3}{15}-\frac{22}{15}=-\frac{19}{15}$$
$$-\frac{19}{15}\cdot\left(-\frac{5}{8}\right)=\frac{19}{24}$$
$$-\frac{5}{12}-2{,}75=-\frac{5}{12}-\frac{11}{4}=-\frac{5}{12}-\frac{33}{12}=-\frac{38}{12}=-\frac{19}{6}$$
$$5\left(-\frac{19}{6}\right):4\frac{1}{3}=5\cdot\left(-\frac{19}{6}\right):\frac{13}{3}=-\frac{95}{6}\cdot\frac{3}{13}=-\frac{95}{26}$$
$$\frac{19}{24}-\left(-\frac{95}{26}\right)=\frac{19}{24}+\frac{95}{26}=\frac{247}{312}+\frac{1140}{312}=\frac{1387}{312}$$

Ответ: а) $$27{,}09$$; б) $$\frac{1387}{312}$$.