Odpowiedź:
[tex]e)\ \ \frac{1}{3}\cdot3^2+\frac{1}{4}\cdot2^2=\frac{1}{\not3_{1}}\cdot\not9^3+\frac{1}{\not4_{1}}\cdot\not4^1=3+1=4\\\\g)\ \ 3\cdot(\frac{1}{3})^3+8\cdot(-\frac{1}{2})^4=\not3^1\cdot\frac{1}{\not27_{9}}+\not8^1\cdot\frac{1}{\not16_{2}}=\frac{1}{9}+\frac{1}{2}=\frac{2}{18}+\frac{9}{18}=\frac{11}{18}\\\\k)\ \ 3^3-(-2)^2-(-1)^6=27-4-1=23-1=22[/tex]
