Odpowiedź:
[tex]c)\ \ \dfrac{\frac{2}{4}+\frac{1}{6}}{\frac{2}{5}-\frac{1}{2}}+(3\frac{2}{18}-4\frac{6}{18})\cdot(0,75-\frac{3}{4})=\dfrac{\frac{1}{2}+\frac{1}{6}}{\frac{4}{10}-\frac{5}{10}}+(-1\frac{4}{18})\cdot(\frac{3}{4}-\frac{3}{4})=\\\\\\=\dfrac{\frac{3}{6}+\frac{1}{6}}{-\frac{1}{10}}-1\frac{4}{18}\cdot0=\dfrac{\frac{4}{6}}{-\frac{1}{10}}-0=\frac{4}{6}:(-\frac{1}{10})=\frac{2}{3}\cdot(-10)=-\frac{20}{3}=-6\frac{2}{3}[/tex]
[tex]d)\ \ \dfrac{5\frac{3}{4}-8\frac{1}{3}}{\frac{1}{2}-(\frac{2}{3})^2}-\frac{2}{3}:(-\frac{4}{27})=\dfrac{\frac{23}{4}-\frac{25}{3}}{\frac{1}{2}-\frac{4}{9}}+\frac{2}{3}:\frac{4}{27}=\dfrac{\frac{69}{12}-\frac{100}{12}}{\frac{9}{18}-\frac{8}{18}}+\frac{\not2^1}{\not3_{1}}\cdot\frac{\not27^9}{\not4_{2}}=\dfrac{-\frac{31}{12}}{\frac{1}{18}}+\frac{9}{2}=\\\\\\=-\frac{31}{12}:\frac{1}{18}+\frac{9}{2}=-\frac{31}{\not12_{2}}\cdot\not18^3+\frac{9}{2}=-\frac{93}{2}+\frac{9}{2}=-\frac{84}{2}=-42[/tex]
