Вопрос:

2. Выполните действие: а) \(4\cdot\left(2\frac{1}{2}+1\frac{3}{4}\right)-\left(6\frac{2}{3}+4\frac{4}{5}\right):2\); б) \(\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:\frac{1}{4}\); в) \(\frac{4}{5}+\frac{4}{493}\cdot\left(7\frac{11}{12}+5\frac{7}{9}\right)\).

Ответ:

а)

\[4\cdot\left(\frac{5}{2}+\frac{7}{4}\right)-\left(\frac{20}{3}+\frac{24}{5}\right):2\]

\[=4\cdot\frac{17}{4}-\frac{172}{15}:2=17-\frac{86}{15}=\frac{169}{15}=11\frac{4}{15}\]

б)

\[\left(6-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\]

\[=\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4=10-\frac{32}{5}=\frac{18}{5}=3\frac{3}{5}\]

в)

\[\frac{4}{5}+\frac{4}{493}\cdot\left(\frac{95}{12}+\frac{52}{9}\right)\]

\[=\frac{4}{5}+\frac{4}{493}\cdot\frac{493}{36}=\frac{4}{5}+\frac{1}{9}=\frac{41}{45}\]

Ответ: а) \(11\frac{4}{15}\); б) \(3\frac{3}{5}\); в) \(\frac{41}{45}\).