Cześć!
Od a) do d)
[tex]\frac{4}{5}+\frac{1}{10}=\frac{4\cdot2}{5\cdot2}+\frac{1}{10}=\frac{8}{10}+\frac{1}{10}=\frac{9}{10}\\\\\frac{1}{6}+\frac{2}{3}=\frac{1}{6}+\frac{2\cdot2}{3\cdot2}=\frac{1}{6}+\frac{4}{6}=\frac{5}{6}\\\\\frac{2}{3}+\frac{1}{5}=\frac{2\cdot5}{3\cdot5}+\frac{1\cdot3}{5\cdot3}=\frac{10}{15}+\frac{3}{15}=\frac{13}{15}\\\\\frac{4}{7}+\frac{1}{2}=\frac{4\cdot2}{7\cdot2}+\frac{1\cdot7}{2\cdot7}=\frac{8}{14}+\frac{7}{14}=\frac{15}{14}=1\frac{1}{14}[/tex]
Od e) do h)
[tex]\frac{8}{9}-\frac{1}{3}=\frac{8}{9}-\frac{1\cdot3}{3\cdot3}=\frac{8}{9}-\frac{3}{9}=\frac{5}{9}\\\\\frac{3}{10}-\frac{1}{5}=\frac{3}{10}-\frac{1\cdot2}{5\cdot2}=\frac{3}{10}-\frac{2}{10}=\frac{1}{10}\\\\\frac{4}{5}-\frac{2}{3}=\frac{4\cdot3}{5\cdot3}-\frac{2\cdot5}{3\cdot5}=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\\\\\frac{5}{7}-\frac{1}{4}=\frac{5\cdot4}{7\cdot4}-\frac{1\cdot7}{4\cdot7}=\frac{20}{28}-\frac{7}{28}=\frac{13}{28}[/tex]
