Вопрос:

39. Решите уравнение: а) \(x+\frac{2}{7}=\frac{5}{14}\); б) \(x-\frac{1}{4}=\frac{6}{11}\); в) \(\frac{3}{8}+h=\frac{11}{18}\); г) \(\frac{4}{15}-y=\frac{7}{30}\); д) \(\frac{3}{8}+a=\frac{5}{6}\); е) \(\frac{7}{12}-y=\frac{3}{32}\); ж) \(\frac{5}{13}+t=\frac{7}{10}\); з) \(\frac{2}{3}-b=\frac{1}{5}\); и) \(t-\frac{1}{6}=\frac{11}{42}\).

Ответ:

  1. \(x=\frac{5}{14}-\frac{2}{7}=\frac{5}{14}-\frac{4}{14}=\frac{1}{14}\).
  2. \(x=\frac{6}{11}+\frac{1}{4}=\frac{24}{44}+\frac{11}{44}=\frac{35}{44}\).
  3. \(h=\frac{11}{18}-\frac{3}{8}=\frac{44}{72}-\frac{27}{72}=\frac{17}{72}\).
  4. \(y=\frac{4}{15}-\frac{7}{30}=\frac{8}{30}-\frac{7}{30}=\frac{1}{30}\).
  5. \(a=\frac{5}{6}-\frac{3}{8}=\frac{20}{24}-\frac{9}{24}=\frac{11}{24}\).
  6. \(y=\frac{7}{12}-\frac{3}{32}=\frac{56}{96}-\frac{9}{96}=\frac{47}{96}\).
  7. \(t=\frac{7}{10}-\frac{5}{13}=\frac{91}{130}-\frac{50}{130}=\frac{41}{130}\).
  8. \(b=\frac{2}{3}-\frac{1}{5}=\frac{10}{15}-\frac{3}{15}=\frac{7}{15}\).
  9. \(t=\frac{11}{42}+\frac{1}{6}=\frac{11}{42}+\frac{7}{42}=\frac{18}{42}=\frac{3}{7}\).

Ответ: \(x=\frac{1}{14};\ x=\frac{35}{44};\ h=\frac{17}{72};\ y=\frac{1}{30};\ a=\frac{11}{24};\ y=\frac{47}{96};\ t=\frac{41}{130};\ b=\frac{7}{15};\ t=\frac{3}{7}\).