Ответ:
- \(\frac{1}{4}\)
- \(\frac{1}{6}\)
- \(\frac{1}{4}\)
- \(\frac{2}{5}\)
- \(\frac{1}{7}\)
- \(\frac{1}{3}\)
- \(\frac{2}{5}\)
- \(\frac{1}{2}\)
- \(\frac{1}{11}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{5}\)
- \(\frac{1}{20}\)
- \(\frac{1}{4}\)
- \(\frac{1}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{8}\)
- \(\frac{a}{2}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{7}{15}\)
- \(\frac{4}{3}\)
- \(\frac{2}{b}\)
- \(\frac{x}{2}\)
- \(\frac{3x}{2}\)
- \(\frac{3}{c}\)
- \(\frac{4}{abc}\)
- \(\frac{2}{x}\)
- \(\frac{y}{2}\)
- \(\frac{2}{c}\)
- \(-\frac{ab}{2ba}=-\frac{1}{2}\)
- \(-\frac{9}{14y}\)
- \(\frac{ab}{5}\)
- \(\frac{1}{3ab^0}=\frac{1}{3a}\)
- \(\frac{1}{2c}\)
- \(\frac{1}{2ab}\)
- \(2xy^2\)
- \(\frac{2}{5}\)
- \(-\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(2\)
- \(\frac{11x^2}{3}\)
- \(\frac{x^3}{2}\)
- \(\frac{6}{a-3}\)
- \(\frac{b^2}{b-3}\)
- \(\frac{3p}{p+1}\)
- \(\frac{10}{y+3}\)
- \(\frac{2}{3b}\)
- \(\frac{3x}{4y}\)
- \(\frac{b}{3a}\)
- \(\frac{3b}{2a}\)
- \(\frac{a+b}{5}\)
- \(\frac{a+3}{a}\)
- \(\frac{2}{x-6}\)
- \(\frac{y}{x+2}\)
- \(\frac{x-y}{x+y}\)
Сокращение выполнено по основному свойству дроби: общий ненулевой множитель числителя и знаменателя сокращён.
