Odpowiedź:
[tex]Zad.9\\\\a)\ \ -(-\frac{1}{3})^3+(\frac{1}{3})^2=-(-\frac{1}{27})+\frac{1}{9}=\frac{1}{27}+\frac{1}{9}=\frac{1}{27}+\frac{3}{27}=\frac{4}{27}\\\\b)\ \ (\frac{3}{5})^2-\frac{3^2}{5}+\frac{3}{5^2}=\frac{9}{25}-\frac{9}{5}+\frac{3}{25}=\frac{12}{25}-\frac{9}{5}=\frac{12}{25}-\frac{45}{25}=-\frac{33}{45}\\\\c)\ \ 10\cdot(-0,3)^3-0,3\cdot10^3=10\cdot(-0,027)-0,3\cdot1000=-0,27-300=-300,27[/tex]
[tex]d)\ \ 7^2\cdot\frac{1}{7}-(-\frac{1}{7})^0\cdot(-7)^2=\not49^7\cdot\frac{1}{\not7_{1}}-1\cdot49=7-49=-42[/tex]
