Odpowiedź:
[tex]b)\ \ \dfrac{\frac{1}{3}\cdot3^2\cdot(-\frac{1}{3})^2}{-\frac{1}{3}}=\dfrac{\frac{1}{3}\cdot\not9\cdot\frac{1}{\not9}}{-\frac{1}{3}}=\dfrac{\frac{1}{3}}{-\frac{1}{3}}=-\frac{1}{3}:\frac{1}{3}=-\frac{1}{\not3}\cdot\not3=-1\\\\\\c)\ \ \dfrac{0,32\cdot(-0,6-0,3)}{3,6}=\dfrac{0,32\cdot(-\not0,9^1)}{\not3,6_{4}}=\dfrac{-0,32}{4}=-0,08[/tex]
[tex]d)\ \ \dfrac{(5\frac{1}{3}-4,5)\cdot(\frac{1}{10}-1)}{1\frac{1}{8}}=\dfrac{(5\frac{1}{3}-4\frac{1}{2})\cdot(\frac{1}{10}-\frac{10}{10})}{\frac{9}{8}}=\dfrac{(5\frac{2}{6}-4\frac{3}{6})\cdot(-\frac{9}{10})}{\frac{9}{8}}=\\\\\\=\dfrac{(4\frac{8}{6}-4\frac{3}{6})\cdot(-\frac{9}{10})}{\frac{9}{8}}=\dfrac{\frac{\not5^1}{\not6_{2}}\cdot(-\frac{\not9^3}{\not10_{2}})}{\frac{9}{8}}=\dfrac{-\frac{3}{4}}{\frac{9}{8}}=-\frac{3}{4}:\frac{9}{8}=-\frac{\not3^1}{\not4_{1}}\cdot\frac{\not8^2}{\not9_{3}}=-\frac{2}{3}[/tex][tex].[/tex]
