Вычислите: а) \(\left(1\frac{5}{7}-1\right)+\left(6\frac{9}{14}-\frac{1}{8}\right)\); б) \(\left(11\frac{1}{2}-\frac{4}{15}\right)-\left(7-\frac{19}{30}\right)\); в) \(\left(46\frac{2}{9}-25\right)-\left(11\frac{4}{7}-\frac{2}{3}\right)\); г) \(\left(29-6\frac{5}{8}\right)+\left(4-\frac{5}{7}\right)\); д) \(7\frac{5}{18}-3\frac{2}{27}-\frac{11}{54}\); е) \(9\frac{3}{20}-2\frac{3}{25}+4\); ж) \(9\frac{11}{42}-\left(2\frac{3}{14}+1\frac{1}{21}\right)\); з) \(5\frac{11}{12}-2\frac{3}{8}-\frac{7}{48}\).

1. \(1\frac{5}{7}-1=\frac{5}{7}\), а \(6\frac{9}{14}-\frac{1}{8}=6+\frac{36-1}{56}=6\frac{35}{56}=6\frac{5}{8}\). Поэтому \(\frac{5}{7}+6\frac{5}{8}=6\frac{75}{56}=7\frac{19}{56}\).
2. \(11\frac{1}{2}-\frac{4}{15}=11\frac{15-8}{30}=11\frac{7}{30}\), \(7-\frac{19}{30}=6\frac{11}{30}\). Разность равна \(4\frac{26}{30}=4\frac{13}{15}\).
3. \(46\frac{2}{9}-25=21\frac{2}{9}\), \(11\frac{4}{7}-\frac{2}{3}=11+\frac{12-14}{21}=10\frac{19}{21}\). Тогда \(21\frac{2}{9}-10\frac{19}{21}=10\frac{40}{63}\).
4. \(29-6\frac{5}{8}=22\frac{3}{8}\), \(4-\frac{5}{7}=3\frac{2}{7}\). Сумма: \(25+\frac{3}{8}+\frac{2}{7}=25\frac{37}{56}\).
5. \(7\frac{5}{18}-3\frac{2}{27}-\frac{11}{54}=7-3+\frac{15-4-11}{54}=4\).
6. \(9\frac{3}{20}-2\frac{3}{25}+4=11+\frac{15-12}{100}=11\frac{3}{100}\).
7. \(2\frac{3}{14}+1\frac{1}{21}=3+\frac{9+2}{42}=3\frac{11}{42}\). Поэтому \(9\frac{11}{42}-3\frac{11}{42}=6\).
8. \(5\frac{11}{12}-2\frac{3}{8}-\frac{7}{48}=3+\frac{44-18-7}{48}=3\frac{19}{48}\).

Ответ: а) \(7\frac{19}{56}\); б) \(4\frac{13}{15}\); в) \(10\frac{40}{63}\); г) \(25\frac{37}{56}\); д) \(4\); е) \(11\frac{3}{100}\); ж) \(6\); з) \(3\frac{19}{48}\).