Odpowiedź:
[tex]0,9:\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{3}{5}=\dfrac{9}{\not10_{2}}*\not5^1+\dfrac{5}{15}+\dfrac{9}{15}=\dfrac{9}{2}+\dfrac{14}{15}=\dfrac{135}{30}+\dfrac{28}{30}=\dfrac{163}{30}=\\\\=5\dfrac{13}{30}\\\\\\3\dfrac{1}{3}-\dfrac{1}{3}*0,6=3\dfrac{1}{3}-\dfrac{1}{\not3_{1}}*\dfrac{\not3^1}{5}=3\dfrac{1}{3}-\dfrac{1}{5}=3\dfrac{5}{15}-\dfrac{3}{15}=3\dfrac{2}{15}[/tex]
[tex](1:0,3-2\dfrac{1}{3})-\dfrac{5}{6}=(1:\dfrac{3}{10}-2\dfrac{1}{3})-\dfrac{5}{6}=(1*\dfrac{10}{3}-\dfrac{7}{3})-\dfrac{5}{6}=\\\\\\=(\dfrac{10}{3}-\dfrac{7}{3})-\dfrac{5}{6}=1-\dfrac{5}{6}=\dfrac{6}{6}-\dfrac{5}{6}=\dfrac{1}{6}\\\\\\6\dfrac{2}{3}*(0,25+\dfrac{1}{5})=\dfrac{20}{3}*(\dfrac{1}{4}+\dfrac{1}{5})=\dfrac{20}{3}*(\dfrac{5}{20}+\dfrac{4}{20})=\dfrac{\not20^1}{\not3_{1}}*\dfrac{\not9^3}{\not20_{1}}=3[/tex]
[tex]1\dfrac{5}{9}:1\dfrac{1}{3}*0,75=\dfrac{14}{9}:\dfrac{4}{3}*\dfrac{3}{4}=\dfrac{\not14^7}{\not9_{3}}*\dfrac{\not3^1}{\not4_{2}}*\dfrac{3}{4}=\dfrac{7}{\not6_{2}}*\dfrac{\not3^1}{4}=\dfrac{7}{8}[/tex]
