а) \(4\frac{2}{15}-3\frac{3}{10}-2\frac{1}{6}=\frac{62}{15}-\frac{33}{10}-\frac{13}{6}=\frac{124-99-65}{30}=-\frac{40}{30}=-\frac{4}{3}=-1\frac{1}{3}\).
б) \(7\frac{5}{21}-14\frac{1}{7}+6\frac{1}{14}=\frac{152}{21}-\frac{99}{7}+\frac{85}{14}=\frac{304-594+255}{42}=-\frac{35}{42}=-\frac{5}{6}\).
в) \(24\frac{2}{35}-18\frac{5}{14}-5\frac{3}{10}=\frac{842}{35}-\frac{257}{14}-\frac{53}{10}=\frac{3368-3598-742}{140}=-\frac{972}{140}=-\frac{243}{35}=-6\frac{33}{35}\).
г) \(1\frac{2}{9}+2\frac{5}{6}-35\frac{1}{5}=\frac{11}{9}+\frac{17}{6}-\frac{176}{5}=\frac{110+255-3168}{90}=-\frac{2803}{90}=-31\frac{13}{90}\).
Ответ: а) \(-1\frac{1}{3}\); б) \(-\frac{5}{6}\); в) \(-6\frac{33}{35}\); г) \(-31\frac{13}{90}\).