1) $$4 \frac{1}{12} - 1 \frac{1}{6} = 4 + \frac{1}{12} - 1 - \frac{1}{6} = (4 - 1) + (\frac{1}{12} - \frac{1}{6}) = 3 + (\frac{1}{12} - \frac{1 \cdot 2}{6 \cdot 2}) = 3 + (\frac{1}{12} - \frac{2}{12}) = 3 - \frac{1}{12} = 2 \frac{11}{12}$$

2) $$9 \frac{11}{24} - 7 \frac{19}{36} = 9 + \frac{11}{24} - 7 - \frac{19}{36} = (9 - 7) + (\frac{11}{24} - \frac{19}{36}) = 2 + (\frac{11 \cdot 3}{24 \cdot 3} - \frac{19 \cdot 2}{36 \cdot 2}) = 2 + (\frac{33}{72} - \frac{38}{72}) = 2 - \frac{5}{72} = 1 \frac{67}{72}$$

3) $$5 \frac{5}{9} - 2 \frac{4}{7} = 5 + \frac{5}{9} - 2 - \frac{4}{7} = (5 - 2) + (\frac{5}{9} - \frac{4}{7}) = 3 + (\frac{5 \cdot 7}{9 \cdot 7} - \frac{4 \cdot 9}{7 \cdot 9}) = 3 + (\frac{35}{63} - \frac{36}{63}) = 3 - \frac{1}{63} = 2 \frac{62}{63}$$

4) $$9 \frac{8}{35} - 2 \frac{52}{105} = 9 + \frac{8}{35} - 2 - \frac{52}{105} = (9 - 2) + (\frac{8}{35} - \frac{52}{105}) = 7 + (\frac{8 \cdot 3}{35 \cdot 3} - \frac{52}{105}) = 7 + (\frac{24}{105} - \frac{52}{105}) = 7 - \frac{28}{105} = 7 - \frac{4}{15} = 6 \frac{11}{15}$$

5) $$8 \frac{6}{35} - 6 \frac{7}{10} = 8 + \frac{6}{35} - 6 - \frac{7}{10} = (8 - 6) + (\frac{6}{35} - \frac{7}{10}) = 2 + (\frac{6 \cdot 2}{35 \cdot 2} - \frac{7 \cdot 7}{10 \cdot 7}) = 2 + (\frac{12}{70} - \frac{49}{70}) = 2 - \frac{37}{70} = 1 \frac{33}{70}$$

6) $$9 \frac{5}{14} - 3 \frac{27}{49} = 9 + \frac{5}{14} - 3 - \frac{27}{49} = (9 - 3) + (\frac{5}{14} - \frac{27}{49}) = 6 + (\frac{5 \cdot 7}{14 \cdot 7} - \frac{27 \cdot 2}{49 \cdot 2}) = 6 + (\frac{35}{98} - \frac{54}{98}) = 6 - \frac{19}{98} = 5 \frac{79}{98}$$

Ответ: 1) 2 11/12; 2) 1 67/72; 3) 2 62/63; 4) 6 11/15; 5) 1 33/70; 6) 5 79/98

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