e) $$10-2\frac{1}{2}:\frac{3}{4}+ (2\frac{1}{2}-1\frac{1}{3})\cdot 6=$$

e) $$10-2\frac{1}{2}:\frac{3}{4}+ (2\frac{1}{2}-1\frac{1}{3})\cdot 6=10-\frac{5}{2}:\frac{3}{4}+(\frac{5}{2}-\frac{4}{3})\cdot6=10-\frac{5}{2}\cdot\frac{4}{3}+(\frac{5\cdot3}{2\cdot3}-\frac{4\cdot2}{3\cdot2})\cdot6=10-\frac{20}{6}+(\frac{15}{6}-\frac{8}{6})\cdot6=10-\frac{10}{3}+\frac{7}{6}\cdot6=10-\frac{10}{3}+7=17-\frac{10}{3}=\frac{17\cdot3}{1\cdot3}-\frac{10}{3}=\frac{51}{3}-\frac{10}{3}=\frac{41}{3}=13\frac{2}{3}$$

Ответ: $$13\frac{2}{3}$$