Odpowiedź:
[tex]Zad.22\\\\a)\ \ \not4^2\cdot\dfrac{15}{\not26_{13}}\cdot\not2^1\cdot\dfrac{13}{\not40_{20}}=2\cdot\dfrac{15}{13}\cdot\dfrac{13}{20}=\dfrac{\not30^3}{\not13_{1}}\cdot\dfrac{\not13^1}{\not20_{2}}=\dfrac{3}{2}=1\dfrac{1}{2}\\\\\\b)\ \ \not6^1\cdot\dfrac{21}{\not24_{4}}\cdot\dfrac{2}{7}\cdot3=\dfrac{\not21^3}{\not4_{2}}\cdot\dfrac{\not6^3}{\not7_{1}}=\dfrac{9}{2}=4\dfrac{1}{2}[/tex]
[tex]c)\ \ 1\dfrac{1}{5}\cdot2\cdot\dfrac{1}{3}\cdot2\dfrac{1}{4}=\dfrac{\not6^2}{5}\cdot\dfrac{2}{\not3_{1}}\cdot\dfrac{9}{4}=\dfrac{\not4^1}{5}\cdot\dfrac{9}{\not4_{1}}=\dfrac{9}{5}=1\dfrac{4}{5}[/tex]
