\(4\frac{3}{4}\cdot\left(2\frac{1}{5}+\frac{7}{8}-1\frac{1}{2}\right)\)

\(4\frac{3}{4}=\frac{19}{4}\), а \(2\frac{1}{5}+\frac{7}{8}-1\frac{1}{2}=\frac{11}{5}+\frac{7}{8}-\frac{3}{2}=\frac{63}{40}\).

\(\frac{19}{4}\cdot\frac{63}{40}=\frac{1197}{160}=7\frac{77}{160}\).

Ответ: \(7\frac{77}{160}\).