Odpowiedź:
[tex]a)\ \ \not3^1\cdot\frac{4}{\not9_{3}}\cdot12=\frac{4}{\not3_{1}}\cdot\not12^4=4\cdot4=16\\\\\\b)\ \ 9\cdot\frac{17}{\not12_{2}}\cdot\not6^1=\frac{9\cdot17}{2}=\frac{153}{2}=76\frac{1}{2}\\\\\\c)\ \ 4\frac{5}{6}\cdot9\cdot2=\frac{29}{\not6_{1}}\cdot\not18^3=29\cdot3=87\\\\\\d)\ \ 9\cdot(10\frac{11}{18}-3\frac{5}{6})=9\cdot(10\frac{11}{18}-3\frac{15}{18})=9\cdot(9\frac{29}{18}-3\frac{15}{18})=9\cdot6\frac{14}{18}=9\cdot6\frac{7}{9}=\not9\cdot\frac{61}{\not9}=61[/tex]
[tex]e)\ \ 3\cdot(1\frac{3}{4}+2\frac{5}{6})=3\cdot(1\frac{9}{12}+2\frac{10}{12})=3\cdot3\frac{19}{12}=\not3^1\cdot\frac{55}{\not12_{4}}=\frac{55}{4}=13\frac{3}{4}\\\\\\f)\ \ 4\cdot(2\frac{1}{4}-1\frac{3}{5})=4\cdot(\frac{9}{4}-\frac{8}{5})=4\cdot(\frac{45}{20}-\frac{32}{20})=\not4^1\cdot\frac{13}{\not20_{5}}=\frac{13}{5}=2\frac{3}{5}[/tex]
