Odpowiedź:
[tex]a)\ \ 5\frac{1}{3}\cdot1\frac{1}{2}-6:4\frac{2}{3}=\frac{\not16^8}{\not3_{1}}\cdot\frac{\not3^1}{\not2_{1}}-6:\frac{14}{3}=8-\not6^3\cdot\frac{3}{\not14_{7}}=8-\frac{9}{7}=8-1\frac{2}{7}=7\frac{7}{7}-1\frac{2}{7}=6\frac{5}{7}[/tex]
[tex]b)\ \ 3\frac{1}{5}\cdot(2\frac{1}{2}-1\frac{1}{4}:6)=\frac{16}{5}\cdot(2\frac{1}{2}-\frac{5}{4}\cdot\frac{1}{6})=\frac{16}{5}\cdot(\frac{5}{2}-\frac{5}{24})=\frac{16}{5}\cdot(\frac{60}{24}-\frac{5}{24})=\\\\=\frac{\not16^2}{\not5_{1}}\cdot\frac{\not55^1^1}{\not24_{3}}=\frac{22}{3}=7\frac{1}{3}[/tex]
[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{10}{7}\cdot\frac{3}{6}=\frac{\not10^5}{7}\cdot\frac{1}{\not2_{1}}=\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{\frac{30}{7}}{\frac{10}{3}\cdot(\frac{5}{4}-\frac{2}{4})}=\dfrac{\frac{30}{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{12}{7}=1\frac{5}{7}[/tex]
