Odpowiedź:
[tex]e)\ \ \frac{7}{9}:\frac{2}{3}*\frac{6}{7}=\frac{7}{\not9_{3}}*\frac{\not3^1}{2}*\frac{6}{7}=\frac{7}{6}*\frac{6}{7}=1\\\\\\f)\ \ 3\frac{3}{4}*2\frac{2}{5}-4\frac{3}{8}=\frac{\not15^3}{\not4_{1}}*\frac{\not12^3}{\not5_{1}}-4\frac{3}{8}=9-4\frac{3}{8}=8\frac{8}{8}-4\frac{3}{8}=4\frac{5}{8}\\\\\\g)\ \ 10-1\frac{5}{6}*4\frac{1}{2}=10-\frac{11}{\not6_{2}}*\frac{\not9^3}{2}=10-\frac{33}{4}=9\frac{4}{4}-8\frac{1}{4}=1\frac{3}{4}[/tex]
[tex]h)\ \ \frac{5}{7}*4\frac{2}{3}+\frac{5}{6}:1\frac{1}{9}=\frac{5}{\not7_{1}}*\frac{\not14^2}{3}+\frac{5}{6}:\frac{10}{9}=\frac{10}{3}+\frac{\not5^1}{\not6_{2}}*\frac{\not9^3}{\not10_{2}}=3\frac{1}{3}+\frac{3}{4}=\\\\=3\frac{4}{12}+\frac{9}{12}=3\frac{13}{12}=4\frac{1}{12}[/tex]
[tex]i)\ \ (2\frac{1}{4}-1\frac{2}{3}):(\frac{1}{6}+\frac{2}{9})=(\frac{9}{4}-\frac{5}{3}):(\frac{3}{18}+\frac{4}{18})=(\frac{27}{12}-\frac{20}{12}):\frac{7}{18}=\frac{\not7^1}{\not12_{2} }*\frac{\not18^3}{\not7_{1}}=\frac{3}{2}=1\frac{1}{2}[/tex]
