а) $\left(\frac{17}{24}-\frac{1}{12}+\frac{1}{8}\right)\cdot24:5$; б) $\left(\frac{13}{18}+\frac{5}{6}-\frac{7}{9}\right)\cdot6:7$; в) $\left(\frac{27}{56}+\frac{11}{14}-\frac{9}{28}\right)\cdot8:5$; г) $\left(\frac{10}{13}-\frac{21}{52}+\frac{15}{26}\right)\cdot39:7$.

Question

Вопрос: $Фото задания: а) $\left(\frac{17}{24}-\frac{1}{12}+\frac{1}{8}\right)\cdot24:5$; б)…$ Фото задания

Accepted Answer

а) $\left(\frac{17}{24}-\frac{2}{24}+\frac{3}{24}\right)\cdot24:5 = \frac{18}{24}\cdot24:5 = 18:5 = \frac{18}{5}$.

б) $\left(\frac{13}{18}+\frac{15}{18}-\frac{14}{18}\right)\cdot6:7 = \frac{14}{18}\cdot6:7 = \frac{7}{9}\cdot6:7 = \frac{2}{3}$.

в) $\left(\frac{27}{56}+\frac{44}{56}-\frac{18}{56}\right)\cdot8:5 = \frac{53}{56}\cdot8:5 = \frac{53}{35}$.

г) $\left(\frac{40}{52}-\frac{21}{52}+\frac{30}{52}\right)\cdot39:7 = \frac{49}{52}\cdot39:7 = \frac{21}{4}$.

Ответ: а) $\frac{18}{5}$; б) $\frac{2}{3}$; в) $\frac{53}{35}$; г) $\frac{21}{4}$.

а) \(\left(\frac{17}{24}-\frac{1}{12}+\frac{1}{8}\right)\cdot24:5\); б) \(\left(\frac{13}{18}+\frac{5}{6}-\frac{7}{9}\right)\cdot6:7\); в) \(\left(\frac{27}{56}+\frac{11}{14}-\frac{9}{28}\right)\cdot8:5\); г) \(\left(\frac{10}{13}-\frac{21}{52}+\frac{15}{26}\right)\cdot39:7\).