Odpowiedź:
[tex]f)\ \ (2\frac{1}{6}-\frac{2}{3})^2-2\frac{1}{5}=(\frac{13}{6}-\frac{2}{3})^2-2\frac{1}{5}=(\frac{13}{6}-\frac{4}{6})^2-2\frac{1}{5}=(\frac{9}{6})^2-2\frac{1}{5}=(\frac{3}{2})^2-2\frac{1}{5}=\\\\=\frac{9}{4}-\frac{11}{5}=\frac{45}{20}-\frac{44}{20}=\frac{1}{20}[/tex]
[tex]g)\ \ 10-2\cdot(3\frac{1}{5}-2\frac{1}{4}\cdot\frac{1}{3})=10-2\cdot(3\frac{1}{5}-\frac{\not9^3}{4}\cdot\frac{1}{\not3_{1}})=10-2\cdot(\frac{16}{5}-\frac{3}{4})=\\\\=10-2\cdot(\frac{64}{20}-\frac{15}{20})=10-\not2^1\cdot\frac{49}{\not20_{10}}=10-\frac{49}{10}=9\frac{10}{10}-4\frac{9}{10}=5\frac{1}{10}[/tex]
[tex]h)\ \ (\frac{7}{3}-\frac{5}{12}+1\frac{1}{2}):(2\frac{3}{4}-\frac{5}{6})=(\frac{28}{12}-\frac{5}{12}+1\frac{6}{12}):(\frac{11}{4}-\frac{5}{6})=(\frac{23}{12}+\frac{18}{12}):(\frac{33}{12}-\frac{10}{12})=\\\\=\frac{41}{12}:\frac{23}{12}=\frac{41}{\not12}\cdot\frac{\not12}{23}=\frac{41}{23}=1\frac{18}{23}[/tex]
