Cześć!
Od a) do d)
[tex]2\frac{1}{3}=\frac{2\cdot3+1}{3}=\frac{6+1}{3}=\frac{7}{3}\\\\3\frac{1}{2}=\frac{3\cdot2+1}{2}=\frac{6+1}{2}=\frac{7}{2}\\\\2\frac{3}{8}=\frac{2\cdot8+3}{8}=\frac{16+3}{8}=\frac{19}{8}\\\\6\frac{2}{3}=\frac{6\cdot3+2}{3}=\frac{18+2}{3}=\frac{20}{3}[/tex]
Od e) do h)
[tex]3\frac{3}{5}=\frac{3\cdot5+3}{5}=\frac{15+3}{5}=\frac{18}{5}\\\\2\frac{7}{9}=\frac{2\cdot9+7}{9}=\frac{18+7}{9}=\frac{25}{9}\\\\9\frac{3}{4}=\frac{9\cdot4+3}{4}=\frac{36+3}{4}=\frac{39}{4}\\\\10\frac{1}{6}=\frac{10\cdot6+1}{6}=\frac{60+1}{6}=\frac{61}{6}[/tex]
Od i) do l)
[tex]7\frac{5}{7}=\frac{7\cdot7+5}{7}=\frac{49+5}{7}=\frac{54}{7}\\\\2\frac{6}{13}=\frac{2\cdot13+6}{13}=\frac{26+6}{13}=\frac{32}{13}\\\\11\frac{7}{9}=\frac{11\cdot9+7}{9}=\frac{99+7}{9}=\frac{106}{9}\\\\100\frac{1}{4}=\frac{100\cdot4+1}{4}=\frac{400+1}{4}=\frac{401}{4}[/tex]
