Odpowiedź:
[tex]c)\ \ 1\frac{3}{7}\cdot(2\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}:1\frac{1}{7})=\frac{10}{7}\cdot(\frac{5}{2}:\frac{3}{2}-\frac{4}{3}:\frac{8}{7})=\frac{10}{7}\cdot(\frac{5}{\not2_{1}}\cdot\frac{\not2^1}{3}-\frac{\not4^1}{3}\cdot\frac{7}{\not8_{2}})=\\\\=\frac{10}{7}\cdot(\frac{5}{3}-\frac{7}{6})=\frac{10}{7}\cdot(\frac{10}{6}-\frac{7}{6})=\frac{\not10^5}{7}\cdot\frac{3}{\not6_{3}}=\frac{15}{21}=\frac{5}{7}[/tex]
[tex]d)\ \ \dfrac{5\frac{4}{7}-1\frac{2}{7}}{3\frac{1}{3}\cdot(1\frac{1}{4}-\frac{1}{2})}=\dfrac{4\frac{2}{7}}{\frac{10}{3}\cdot(\frac{5}{4}-\frac{1}{2})}=\dfrac{4\frac{2}{7}}{\frac{10}{3}\cdot(\frac{5}{4}-\frac{2}{4})}=\dfrac{4\frac{2}{7}}{\frac{\not10^5}{\not3_{1}}\cdot\frac{\not3^1}{\not4_{2}}}=\dfrac{\frac{30}{7}}{\frac{5}{2}}=\frac{30}{7}:\frac{5}{2}=\\\\\\=\frac{\not30^6}{7}\cdot\frac{2}{\not5_{1}}=\frac{x}{y}=\frac{12}{7}=1\frac{5}{7}[/tex]
