2.287. Вычислите: а) $\frac{7}{16}\cdot\frac{4}{9}-\frac{2}{15}$; б) $\frac{8}{15}\cdot\left(2\frac{1}{2}\right)^2-\frac{5}{9}$; в) $\left(1\frac{1}{6}\right)^2-\frac{7}{18}\cdot2\frac{4}{7}-1\frac{2}{5}$; г) $\left(\frac{11}{16}+\frac{7}{24}-\frac{5}{12}\right)+\left(\frac{3}{4}\right)^2$.

Question

2.287. Вычислите: а) $\frac{7}{16}\cdot\frac{4}{9}-\frac{2}{15}$; б) $\frac{8}{15}\cdot\left(2\frac{1}{2}\right)^2-\frac{5}{9}$; в) $\left(1\frac{1}{6}\right)^2-\frac{7}{18}\cdot2\frac{4}{7}-1\frac{2}{5}$; г) $\left(\frac{11}{16}+\frac{7}{24}-\frac{5}{12}\right)+\left(\frac{3}{4}\right)^2$.

$Фото задания: 2.287. Вычислите: а) $\frac{7}{16}\cdot\frac{4}{9}-\frac{2}{15}$; б)…$ Фото задания

Accepted Answer

а) $\frac{7}{16}\cdot\frac{4}{9}-\frac{2}{15}=\frac{7}{36}-\frac{2}{15}=\frac{35-24}{180}=\frac{11}{180}$.

б) $\frac{8}{15}\cdot\left(2\frac{1}{2}\right)^2-\frac{5}{9}=\frac{8}{15}\cdot\left(\frac{5}{2}\right)^2-\frac{5}{9}=\frac{8}{15}\cdot\frac{25}{4}-\frac{5}{9}=\frac{10}{3}-\frac{5}{9}=\frac{25}{9}=2\frac{7}{9}$.

в) $\left(1\frac{1}{6}\right)^2-\frac{7}{18}\cdot2\frac{4}{7}-1\frac{2}{5}=\left(\frac{7}{6}\right)^2-\frac{7}{18}\cdot\frac{18}{7}-\frac{7}{5}=\frac{49}{36}-1-\frac{7}{5}=\frac{13}{36}-\frac{7}{5}=\frac{65-252}{180}=-\frac{187}{180}=-1\frac{7}{180}$.

г) $\left(\frac{11}{16}+\frac{7}{24}-\frac{5}{12}\right)+\left(\frac{3}{4}\right)^2=\left(\frac{33}{48}+\frac{14}{48}-\frac{20}{48}\right)+\frac{9}{16}=\frac{27}{48}+\frac{27}{48}=\frac{54}{48}=\frac{9}{8}=1\frac{1}{8}$.

Ответ: а) $\frac{11}{180}$; б) $2\frac{7}{9}$; в) $-1\frac{7}{180}$; г) $1\frac{1}{8}$.