Ответ:

Чтобы привести дроби к общему знаменателю, найдём наименьшее общее кратное знаменателей и домножим числитель и знаменатель каждой дроби на соответствующий множитель.

  1. а) НОЗ(6, 8) = 24: \(\frac{1}{6}=\frac{4}{24}\), \(\frac{3}{8}=\frac{9}{24}\).
  2. б) НОЗ(9, 15) = 45: \(\frac{4}{9}=\frac{20}{45}\), \(\frac{7}{15}=\frac{21}{45}\).
  3. в) НОЗ(12, 8) = 24: \(\frac{5}{12}=\frac{10}{24}\), \(\frac{1}{8}=\frac{3}{24}\).
  4. г) НОЗ(15, 12) = 60: \(\frac{8}{15}=\frac{32}{60}\), \(\frac{11}{12}=\frac{55}{60}\).
  5. д) НОЗ(10, 12) = 60: \(\frac{9}{10}=\frac{54}{60}\), \(\frac{5}{12}=\frac{25}{60}\).
  6. е) НОЗ(12, 18) = 36: \(\frac{13}{12}=\frac{39}{36}\), \(\frac{13}{18}=\frac{26}{36}\).
  7. ж) НОЗ(30, 45) = 90: \(\frac{11}{30}=\frac{33}{90}\), \(\frac{8}{45}=\frac{16}{90}\).
  8. з) НОЗ(20, 16) = 80: \(\frac{11}{20}=\frac{44}{80}\), \(\frac{9}{16}=\frac{45}{80}\).
  9. и) НОЗ(33, 77) = 231: \(\frac{8}{33}=\frac{56}{231}\), \(\frac{9}{77}=\frac{27}{231}\).
  10. к) НОЗ(98, 56) = 392: \(\frac{9}{98}=\frac{36}{392}\), \(\frac{5}{56}=\frac{35}{392}\).
  11. л) НОЗ(750, 456) = 57000: \(\frac{13}{750}=\frac{988}{57000}\), \(\frac{7}{456}=\frac{875}{57000}\).
  12. м) НОЗ(297, 363) = 3267: \(\frac{10}{297}=\frac{110}{3267}\), \(\frac{14}{363}=\frac{126}{3267}\).

Ответ: а) \(\frac{4}{24}\) и \(\frac{9}{24}\); б) \(\frac{20}{45}\) и \(\frac{21}{45}\); в) \(\frac{10}{24}\) и \(\frac{3}{24}\); г) \(\frac{32}{60}\) и \(\frac{55}{60}\); д) \(\frac{54}{60}\) и \(\frac{25}{60}\); е) \(\frac{39}{36}\) и \(\frac{26}{36}\); ж) \(\frac{33}{90}\) и \(\frac{16}{90}\); з) \(\frac{44}{80}\) и \(\frac{45}{80}\); и) \(\frac{56}{231}\) и \(\frac{27}{231}\); к) \(\frac{36}{392}\) и \(\frac{35}{392}\); л) \(\frac{988}{57000}\) и \(\frac{875}{57000}\); м) \(\frac{110}{3267}\) и \(\frac{126}{3267}\).