Odpowiedź:
[tex]a)\ \ \frac{2}{3}+\frac{7}{9}-\frac{5}{6}=\frac{12}{18}+\frac{14}{18}-\frac{15}{18}=\frac{11}{18}\\\\\\b)\ \ 1\frac{1}{2}-\frac{3}{4}+\frac{5}{8}=\frac{3}{2}-\frac{3}{4}+\frac{5}{8}=\frac{12}{8}-\frac{6}{8}+\frac{5}{8}=\frac{11}{8}=1\frac{3}{8}\\\\\\c)\ \ \frac{\not3^1}{4}\cdot\frac{5}{\not6_{2}}:\frac{1}{2}=\frac{5}{\not8_{4}}\cdot\not2^1=\frac{5}{4}=1\frac{1}{4}[/tex]
[tex]d)\ \ \frac{7}{9}:\frac{2}{3}\cdot\frac{6}{7}=\frac{7}{\not9_{3}}\cdot\frac{\not3^1}{2}\cdot\frac{6}{7}=\frac{\not7^1}{\not6_{1}}\cdot\frac{\not6^1}{\not7_{1}}=1\\\\\\e)\ \ 3\frac{3}{4}\cdot2\frac{2}{5}-4\frac{3}{8}=\frac{\not15^3}{\not4_{1}}\cdot\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}[/tex]
[tex]f)\ \ 10-1\frac{5}{6}\cdot4\frac{1}{2}=10-\frac{11}{\not6_{2}}\cdot\frac{\not9^3}{2}=10-\frac{33}{4}=10-8\frac{1}{4}=9\frac{4}{4}-8\frac{1}{4}=1\frac{3}{4}[/tex]
