Упр.18.16 Часть 1 ГДЗ Мордкович Семенов 7 класс (Алгебра)

Задача

Вычислите: а) $$\left(1{,}5-\frac{12}{35}:\frac{2}{5}\right)\cdot\left(1\frac{13}{21}+1\frac{5}{7}\right)$$; б) $$\left(1\frac{5}{6}\cdot3\frac{3}{4}-4\frac{13}{24}\right):\frac{5}{12}+1\frac{24}{35}$$.

Подробный ответ

а)

$$(1{,}5-\frac{12}{35}:\frac{2}{5})\cdot\left(1\frac{13}{21}+1\frac{5}{7}\right)$$
$$1{,}5=\frac{3}{2}, \qquad \frac{12}{35}:\frac{2}{5}=\frac{12}{35}\cdot\frac{5}{2}=\frac{6}{7}$$
$$\frac{3}{2}-\frac{6}{7}=\frac{21-12}{14}=\frac{9}{14}$$
$$1\frac{13}{21}+1\frac{5}{7}=2+\frac{13}{21}+\frac{15}{21}=2+\frac{28}{21}=2+\frac{4}{3}=\frac{10}{3}$$
$$\frac{9}{14}\cdot\frac{10}{3}=\frac{15}{7}=2\frac{1}{7}$$

б)

$$\left(1\frac{5}{6}\cdot 3\frac{3}{4}-4\frac{13}{24}\right):\frac{5}{12}+1\frac{24}{35}$$
$$1\frac{5}{6}=\frac{11}{6}, \qquad 3\frac{3}{4}=\frac{15}{4}$$
$$\frac{11}{6}\cdot\frac{15}{4}=\frac{55}{8}$$
$$\frac{55}{8}-4\frac{13}{24}=\frac{55}{8}-\frac{109}{24}=\frac{165-109}{24}=\frac{56}{24}=\frac{7}{3}$$
$$\frac{7}{3}:\frac{5}{12}=\frac{7}{3}\cdot\frac{12}{5}=\frac{28}{5}$$
$$1\frac{24}{35}=\frac{59}{35}$$
$$\frac{28}{5}+\frac{59}{35}=\frac{196}{35}+\frac{59}{35}=\frac{255}{35}=\frac{51}{7}=7\frac{2}{7}$$