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

Задача

а) (1 1/4+1 2/3-7/12) :1 1/6;
б) 2 4/15-4 1/5•(1 5/6-41/42);
в) (2/15+1 7/12)•30/103-2:2 1/4•9/32;
г) (3 1/2 :4 2/3+4 2/3 :3 1/2)•4 4/5.

Подробный ответ

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)

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

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

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