Решение:
- \( \frac{8}{15} + \frac{7}{12} = \frac{8 \cdot 4}{15 \cdot 4} + \frac{7 \cdot 5}{12 \cdot 5} = \frac{32}{60} + \frac{35}{60} = \frac{32+35}{60} = \frac{67}{60} \)
- \( \frac{21}{50} - \frac{7}{30} = \frac{21 \cdot 3}{50 \cdot 3} - \frac{7 \cdot 5}{30 \cdot 5} = \frac{63}{150} - \frac{35}{150} = \frac{63-35}{150} = \frac{28}{150} = \frac{14}{75} \)
- \( \frac{5}{42} + \frac{5}{36} = \frac{5 \cdot 6}{42 \cdot 6} + \frac{5 \cdot 7}{36 \cdot 7} = \frac{30}{252} + \frac{35}{252} = \frac{30+35}{252} = \frac{65}{252} \)
- \( \frac{11}{24} + \frac{4}{15} = \frac{11 \cdot 5}{24 \cdot 5} + \frac{4 \cdot 8}{15 \cdot 8} = \frac{55}{120} + \frac{32}{120} = \frac{55+32}{120} = \frac{87}{120} = \frac{29}{40} \)
- \( \frac{11}{30} - \frac{7}{80} = \frac{11 \cdot 8}{30 \cdot 8} - \frac{7 \cdot 3}{80 \cdot 3} = \frac{88}{240} - \frac{21}{240} = \frac{88-21}{240} = \frac{67}{240} \)
- \( \frac{19}{35} + \frac{10}{21} = \frac{19 \cdot 3}{35 \cdot 3} + \frac{10 \cdot 5}{21 \cdot 5} = \frac{57}{105} + \frac{50}{105} = \frac{57+50}{105} = \frac{107}{105} \)
Ответ: a) \( \frac{67}{60} \), \( \frac{14}{75} \), \( \frac{65}{252} \); б) \( \frac{29}{40} \), \( \frac{67}{240} \), \( \frac{107}{105} \).