Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №3796133
А
Аноним
10 часов назад · публичный вопрос
А
Аноним

Реши с решением

Ответ нейросети
10 часов назад

Ответы:

  1. \(4+\frac{6}{73}=4\frac{6}{73}\)
  2. \(5+2\frac{3}{14}=7\frac{3}{14}\)
  3. \(7\frac{3}{19}+\frac{16}{19}=8\)
  4. \(12\frac{2}{3}+3=15\frac{2}{3}\)
  5. \(5\frac{8}{9}-3=2\frac{8}{9}\)
  6. \(6-\frac{3}{19}=5\frac{16}{19}\)
  7. \(7\frac{3}{8}-2\frac{1}{8}=5\frac{1}{4}\)
  8. \(13\frac{7}{17}-9\frac{13}{17}=3\frac{11}{17}\)
  9. \(1\frac{2}{25}-\frac{11}{25}=\frac{16}{25}\)
  10. \(9\frac{7}{12}+\frac{1}{12}=9\frac{2}{3}\)
  11. \(7\frac{1}{8}+\frac{3}{8}=7\frac{1}{2}\)
  12. \(17\frac{9}{10}+50\frac{7}{10}=68\frac{3}{5}\)
  13. \(7\frac{5}{6}-1=6\frac{5}{6}\)
  14. \(10\frac{5}{14}-8\frac{3}{14}=2\frac{1}{7}\)
  15. \(12\frac{3}{10}-\frac{9}{10}=11\frac{2}{5}\)
  16. \(1-\frac{11}{14}=\frac{3}{14}\)
  17. \(8\frac{10}{51}+\frac{7}{51}=8\frac{1}{3}\)
  18. \(\frac{21}{32}+\frac{19}{32}=1\frac{1}{4}\)
  19. \(8\frac{10}{51}+\frac{7}{51}=8\frac{1}{3}\)
  20. \(5\frac{4}{15}+13=18\frac{4}{15}\)
  21. \(12\frac{7}{11}-\frac{10}{11}=11\frac{8}{11}\)
  22. \(9\frac{1}{33}-\frac{4}{33}=8\frac{10}{11}\)
  23. \(8-3\frac{1}{12}=4\frac{11}{12}\)
  24. \(1-\frac{10}{23}=\frac{13}{23}\)
  25. \(\frac{27}{28}+4\frac{9}{28}=5\frac{2}{7}\)
  26. \(3\frac{10}{13}+5\frac{11}{13}=9\frac{8}{13}\)
  27. \(7\frac{3}{10}+3\frac{7}{10}=11\)
  28. \(31\frac{1}{9}+12\frac{8}{9}=44\)