Упр.572 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

1. Найдите значение выражения: а) $$\left(1\frac{1}{4}+1\frac{2}{3}-\frac{7}{12}\right)\div1\frac{1}{6}$$; б) $$2\frac{4}{15}-4\frac{1}{5}\cdot\left(1\frac{5}{6}-\frac{41}{42}\right)$$; в) $$\left(\frac{2}{15}+1\frac{7}{12}\right)\cdot\frac{30}{103}-2\div2\frac{1}{4}\cdot\frac{9}{32}$$; г) $$\left(3\frac{1}{2}\div4\frac{2}{3}+4\frac{2}{3}\div3\frac{1}{2}\right)\cdot4\frac{4}{5}$$.

а) Преобразуем смешанные числа в неправильные дроби:

$$\left(1\frac{1}{4}+1\frac{2}{3}-\frac{7}{12}\right):1\frac{1}{6} = \left(\frac{5}{4}+\frac{5}{3}-\frac{7}{12}\right):\frac{7}{6}$$

Приведём к общему знаменателю:
$$\frac{5}{4}+\frac{5}{3}-\frac{7}{12} = \frac{15}{12}+\frac{20}{12}-\frac{7}{12} = \frac{28}{12} = \frac{7}{3}$$

Тогда
$$\frac{7}{3}:\frac{7}{6} = \frac{7}{3}\cdot\frac{6}{7} = 2$$

б)

$$2\frac{4}{15}-4\frac{1}{5}\cdot\left(1\frac{5}{6}-\frac{41}{42}\right) = \frac{34}{15}-\frac{21}{5}\cdot\left(\frac{11}{6}-\frac{41}{42}\right)$$

$$\frac{11}{6}-\frac{41}{42} = \frac{77}{42}-\frac{41}{42} = \frac{36}{42} = \frac{6}{7}$$

Тогда
$$\frac{34}{15}-\frac{21}{5}\cdot\frac{6}{7} = \frac{34}{15}-\frac{18}{5} = \frac{34}{15}-\frac{54}{15} = -\frac{20}{15} = -\frac{4}{3} = -1\frac{1}{3}$$

в)

$$\left(\frac{2}{15}+1\frac{7}{12}\right)\cdot\frac{30}{103}-2:2\frac{1}{4}\cdot\frac{9}{32} = \left(\frac{2}{15}+\frac{19}{12}\right)\cdot\frac{30}{103}-2:\frac{9}{4}\cdot\frac{9}{32}$$

$$\frac{2}{15}+\frac{19}{12} = \frac{8}{60}+\frac{95}{60} = \frac{103}{60}$$

$$\frac{103}{60}\cdot\frac{30}{103} = \frac{1}{2}$$

Далее:
$$2:\frac{9}{4}\cdot\frac{9}{32} = 2\cdot\frac{4}{9}\cdot\frac{9}{32} = \frac{1}{4}$$

Тогда
$$\frac{1}{2}-\frac{1}{4} = \frac{1}{4}$$

г)

$$\left(3\frac{1}{2}:4\frac{2}{3}+4\frac{2}{3}:3\frac{1}{2}\right)\cdot4\frac{4}{5} = \left(\frac{7}{2}:\frac{14}{3}+\frac{14}{3}:\frac{7}{2}\right)\cdot\frac{24}{5}$$

$$\frac{7}{2}:\frac{14}{3} = \frac{7}{2}\cdot\frac{3}{14} = \frac{3}{4}, \qquad \frac{14}{3}:\frac{7}{2} = \frac{14}{3}\cdot\frac{2}{7} = \frac{4}{3}$$

$$\left(\frac{3}{4}+\frac{4}{3}\right)\cdot\frac{24}{5} = \frac{25}{12}\cdot\frac{24}{5} = 5\cdot2 = 10$$

Ответ

а) 2; б) $$-1\frac{1}{3}$$; в) $$\frac{1}{4}$$; г) 10.