Odpowiedź:
[tex]c)\ \ -\frac{1}{5}-1\frac{3}{7}\cdot(\frac{7}{12}-2\frac{1}{3}):(-1\frac{2}{3}+3\frac{3}{4})=-\frac{1}{5}-\frac{10}{7}\cdot(\frac{7}{12}-\frac{7}{3}):(-1\frac{8}{12}+3\frac{9}{12})=\\\\=-\frac{1}{5}-\frac{10}{7}\cdot(\frac{7}{12}-\frac{28}{12}):2\frac{1}{12}=-\frac{1}{5}-\frac{\not10^5}{\not7_{1}}\cdot(-\frac{\not21^3}{\not12_{6}}):\frac{25}{12}=-\frac{1}{5}+\frac{\not15^3}{\not6_{1}}\cdot\frac{\not12^2}{\not25_{5}}=-\frac{1}{5}+\frac{6}{5}=\\\\=\frac{5}{5}=1[/tex]
[tex]d)\ \ 5\frac{3}{7}+(-3\frac{1}{3})^2\cdot(-1\frac{2}{25})+4\frac{1}{6}:1\frac{2}{3}=5\frac{3}{7}+(-\frac{10}{3})^2\cdot(-\frac{27}{25})+\frac{25}{6}:\frac{5}{3}=\\\\=5\frac{3}{7}+\frac{\not100^4}{\not9_{1}}\cdot(-\frac{\not27^3}{\not25_{1}})+\frac{\not25^5}{\not6_{2}}\cdot\frac{\not3^1}{\not5_{1}}=5\frac{3}{7}-12+\frac{5}{2}=5\frac{3}{7}+2\frac{1}{2}-12=5\frac{6}{14}+2\frac{7}{14}-12=\\\\=7\frac{13}{14}-12=7\frac{13}{14}-11\frac{14}{14}=-4\frac{1}{14}[/tex]
