Ответ: смотри решение ниже
п)
\[2\frac{5}{7}+3\frac{2}{4}=2\frac{20}{28}+3\frac{14}{28}=5\frac{34}{28}=6\frac{6}{28}=6\frac{3}{14}\]
\[2\frac{3}{4}\cdot2\frac{5}{33}=\frac{11}{4}\cdot\frac{71}{33}=\frac{11\cdot71}{4\cdot33}=\frac{1\cdot71}{4\cdot3}=\frac{71}{12}=5\frac{11}{12}\]
р)
\[-4\frac{2}{3}+7\frac{1}{2}=-4\frac{4}{6}+7\frac{3}{6}=2\frac{5}{6}\]
ш)
\[-3\frac{1}{2}\cdot2\frac{2}{5}=-\frac{7}{2}\cdot\frac{12}{5}=-\frac{7\cdot12}{2\cdot5}=-\frac{7\cdot6}{1\cdot5}=-\frac{42}{5}=-8\frac{2}{5}\]
Ответ:
\[2\frac{5}{7}+3\frac{2}{4}=6\frac{3}{14}\]
\[2\frac{3}{4}\cdot2\frac{5}{33}=5\frac{11}{12}\]
\[-4\frac{2}{3}+7\frac{1}{2}=2\frac{5}{6}\]
\[-3\frac{1}{2}\cdot2\frac{2}{5}=-8\frac{2}{5}\]