Od a) do d)
[tex]\frac{20}{21}-\frac{1}{3}=\frac{20}{21}-\frac{7}{21}=\frac{13}{21}\\\\\frac{5}{12}+2\frac{3}{8}=\frac{10}{24}+2\frac{9}{24}=2\frac{19}{24}\\\\\frac{\not25^5}{\not3_1}\cdot\frac{\not9^3}{\not35_7}=\frac{15}{7}=2\frac{1}{7}\\\\\frac{5}{16}:2\frac{1}{12}=\frac{5}{16}:\frac{25}{12}=\frac{\not5^1}{\not16_4}\cdot\frac{\not12^3}{\not25_5}=\frac{3}{20}[/tex]
Od e) do h)
[tex]\frac{5}{6}+(-0,75)=\frac{5}{6}-\frac{75}{100}=\frac{5}{6}-\frac{3}{4}=\frac{10}{12}-\frac{9}{12}=\frac{1}{12}\\\\3\frac{1}{25}-0,44=3\frac{4}{100}-0,44=3,04-0,44=3-0,4=2,6\\\\-\frac{5}{6}\cdot(-2,4)=-\frac{5}{6}\cdot(-2\frac{4}{10})=-\frac{5}{6}\cdot(-2\frac{2}{5})=-\frac{\not5^1}{\not6_1}\cdot(-\frac{\not12^2}{\not5_1})=2\\\\-0,15:(-\frac{9}{16})=-\frac{15}{100}\cdot(-\frac{16}{9})=-\frac{\not3^1}{\not20_5}\cdot(-\frac{\not16^4}{\not9_3})=\frac{4}{15}[/tex]
