Ответ:
1) Вынесем общий множитель \(5\frac{1}{3}\):
\(5\frac{1}{3}\cdot9-2\frac{3}{4}\cdot5\frac{1}{3}-7\frac{1}{2}\cdot1\frac{5}{9}\)
\(=\frac{16}{3}\cdot9-\frac{11}{4}\cdot\frac{16}{3}-\frac{15}{2}\cdot\frac{14}{9}\)
\(=48-\frac{44}{3}-\frac{35}{3}=48-\frac{79}{3}=\frac{65}{3}=21\frac{2}{3}\).
2) \(2\frac{1}{4}\cdot2\frac{2}{27}+\left(3\frac{1}{6}+4\frac{5}{6}\cdot\frac{11}{29}\right)\cdot1\frac{1}{15}\)
\(=\frac{9}{4}\cdot\frac{56}{27}+\left(\frac{19}{6}+\frac{29}{6}\cdot\frac{11}{29}\right)\cdot\frac{16}{15}\)
\(=\frac{14}{3}+\left(\frac{19}{6}+\frac{11}{6}\right)\cdot\frac{16}{15}\)
\(=\frac{14}{3}+5\cdot\frac{16}{15}=\frac{14}{3}+\frac{16}{3}=10\).
Ответ: 1) \(21\frac{2}{3}\); 2) \(10\).
