Odpowiedź:
[tex]Zad.4\\\\a)\ \ \sqrt[3]{\frac{1}{27}\cdot(-64)}=\sqrt[3]{-\frac{64}{27}}=-\frac{4}{3}=-1\frac{1}{3}\\\\b)\ \ \sqrt[3]{125\cdot\frac{8}{27}\cdot(-0,001)}=\sqrt[3]{125}\cdot\sqrt[3]{\frac{8}{27}}\cdot\sqrt[3]{-0,001}=5\cdot\frac{2}{3}\cdot(-0,1)=\\\\=\frac{\not10}{3}\cdot(-\frac{1}{\not10})=-\frac{1}{3}[/tex]
[tex]Zad.6\\\\\dfrac{9^{-\frac{1}{3}}\cdot\sqrt[6]{3}}{\sqrt{3^{-9}}\cdot9^{\frac{2}{3}}\cdot\sqrt[3]{27^2}}=\dfrac{(3^2)^{-\frac{1}{3}}\cdot3^{\frac{1}{6}}}{3^{-\frac{9}{2}}\cdot(3^2)^{\frac{2}{3}}\cdot27^{\frac{2}{3}}}=\dfrac{3^{-\frac{2}{3}}\cdot3^{\frac{1}{6}}}{3^{-\frac{9}{2}}\cdot3^\frac{4}{3}\cdot(3^3)^{\frac{2}{3}}}=\dfrac{3^{-\frac{2}{3}+\frac{1}{6}}}{3^{-\frac{9}{2}}\cdot3^\frac{4}{3}\cdot3^2}}=[/tex]
[tex]=\dfrac{3^{-\frac{4}{6}+\frac{1}{6}}}{3^{-\frac{9}{2}+\frac{4}{3}+2}}=\dfrac{3^{-\frac{3}{6}}}{3^{-\frac{27}{6}+\frac{8}{6}+\frac{12}{6}}}=\dfrac{3^{-\frac{1}{2}}}{3^{-\frac{7}{6}}}=3^{-\frac{1}{2}-(-\frac{7}{6})}=3^{-\frac{1}{2}+\frac{7}{6}}=3^{-\frac{3}{6}+\frac{7}{6}}=3^{\frac{4}{6}}=3^{\frac{2}{3}}[/tex]
