Odpowiedź:
[tex]e)\ \ (-\frac{1}{4})^2-(-1,8):(-1\frac{1}{8})+1,6\cdot1\frac{2}{3}-(-\frac{3}{4^2})=\frac{1}{16}+\frac{18}{10}:(-\frac{9}{8})+\frac{16}{10}\cdot\frac{5}{3}-(-\frac{3}{16})=\\\\=\frac{1}{16}+\frac{\not9^1}{5}\cdot(-\frac{8}{\not9_{1}})+\frac{8}{\not5_{1}}\cdot\frac{\not5^1}{3}+\frac{3}{16}=\frac{4}{16}-\frac{8}{5}+\frac{8}{3}=\frac{1}{4}-\frac{8}{5}+\frac{8}{3}=\frac{15}{60}-\frac{96}{60}+\frac{160}{60}=\\\\=\frac{79}{60}=1\frac{19}{60}[/tex]
[tex]f)\ \ 1,08:(-0,4)-2\frac{5}{6}:1,7-2\frac{1}{2}\cdot(1,4-1\frac{4}{5})-(-\frac{7}{15})=\\\\=-2,7-\frac{17}{6}:\frac{17}{10}-\frac{5}{2}\cdot(1,4-1,8)+\frac{7}{15}=-2,7-\frac{\not17^1}{\not6_{3}}\cdot\frac{\not10^5}{\not17_{1}}-\frac{5}{2}\cdot(-0,4)+\frac{7}{15}=\\\\=-2,7-\frac{5}{3}-\frac{\not5^1}{\not2_{1}}\cdot(-\frac{\not2^1}{\not5_{1}})+\frac{7}{15}=-2\frac{7}{10}-1\frac{2}{3}+1+\frac{7}{15}=-2\frac{21}{30}-1\frac{20}{30}+1+\frac{14}{30}=[/tex]
[tex]=-3\frac{41}{30}+1\frac{14}{30}=-2\frac{27}{30}=-2\frac{9}{10}[/tex]
