Ответ:Заполняем пропуски:1 \(\frac{3}{4}\) + \(\frac{3}{12}\) = \(\frac{12}{12}\) + \(\frac{9}{12}\) + \(\frac{3}{12}\) = \(\frac{12+9+3}{12}\) = \(\frac{24}{12}\) = 2\(\frac{1}{2}\) - \(\frac{1}{3}\) = \(\frac{3}{6}\) - \(\frac{2}{6}\) = \(\frac{3-2}{6}\) = \(\frac{1}{6}\)