Odpowiedź:
[tex]a)\ \ -1\frac{1}{4}\cdot(-4)=\frac{5}{\not4}\cdot\not4=5\\\\b)\ \ -6\frac{2}{3}:2=-\frac{\not20^1^0}{3}\cdot\frac{1}{\not2_{1}}=-\frac{10}{3}=-3\frac{1}{3}\\\\c)\ \ -0,2\cdot(-0,2)=0,04\\\\d)\ \ -3\frac{1}{3}\cdot(-3)=\frac{10}{\not3}\cdot\not3=10\\\\e)\ \ (-3)^2\cdot\frac{1}{3}=\not9^3\cdot\frac{1}{\not3_{1}}=3\\\\f)\ \ (\frac{1}{3})^2:\frac{1}{2}=\frac{1}{9}\cdot2=\frac{2}{9}[/tex]
[tex]g)\ \ -6^2:\frac{1}{2}=-36\cdot2=-72\\\\h)\ \ 3\frac{3}{4}:(-1,25)^2=\frac{15}{4}:(-\frac{125}{100})^2=\frac{15}{4}:(-\frac{5}{4})^2=\frac{15}{4}:\frac{25}{16}=\frac{\not15^3}{\not4_{1}}\cdot\frac{\not16^4}{\not25_{5}}=\frac{12}{5}=2\frac{2}{5}[/tex]
