Вычисли: \(17\frac34-(6\frac1{32}+5)+4\frac38-(5\frac3{16}+8-11\frac14)+9\frac58\).

Приведём дробные части к знаменателю 32:

\[17\frac34=17\frac{24}{32},\quad 4\frac38=4\frac{12}{32},\quad 9\frac58=9\frac{20}{32},\quad 5\frac3{16}=5\frac6{32},\quad 11\frac14=11\frac8{32}.\]

\[17\frac{24}{32}-\left(6\frac1{32}+5\right)+4\frac{12}{32}-\left(5\frac6{32}+8-11\frac8{32}\right)+9\frac{20}{32}=\frac{561}{32}-\frac{353}{32}+\frac{140}{32}-\frac{58}{32}+\frac{308}{32}=\frac{598}{32}=18\frac{11}{16}.\]

Ответ: \(18\frac{11}{16}\).