Cześć!
Od a) do c)
[tex]3\frac{7}{9}+\frac{6}{9}=3\frac{13}{9}=4\frac{4}{9}\\\\\frac{2}{5}+\frac{1}{3}=\frac{2\cdot3}{5\cdot3}+\frac{1\cdot5}{3\cdot5}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}\\\\\frac{3}{4}-\frac{2}{3}=\frac{3\cdot3}{4\cdot3}-\frac{2\cdot4}{3\cdot4}=\frac{9}{12}-\frac{8}{12}=\frac{1}{12}[/tex]
Od d) do f)
[tex]7\frac{1}{4}-5\frac{3}{4}=6\frac{5}{4}-5\frac{3}{4}=1\frac{2}{4}=1\frac{1}{2}\\\\4\frac{1}{6}+2\frac{7}{9}=4\frac{1\cdot3}{6\cdot3}+2\frac{7\cdot2}{9\cdot2}=4\frac{3}{18}+2\frac{14}{18}=6\frac{17}{18}\\\\5\frac{1}{4}-4\frac{7}{10}=5\frac{1\cdot5}{4\cdot5}-4\frac{7\cdot2}{10\cdot2}=5\frac{5}{20}-4\frac{14}{20}=4\frac{25}{20}-4\frac{14}{20}=\frac{11}{20}[/tex]
