a)
0,125^3*32^{-2}=((0,5)^3)^3*(2^5)^{-2}=(\frac{1}{2})^9*2^{-10}=2^{-9+(-10)}=2^{-19}
b)
(\frac{1}{27})^5:81^{3,25}=((\frac{1}{3})^3)^5:(3^4)^{3,25}=(\frac{1}{3})^{15}:3^{13}=3^{-15}:3^{13}=3^{-15-13}=3^{-28}
c)
{0,4^7*2,5^{-4}:(\frac{2}{5})^5=0,4^7*\frac{2,5^{-4}}{0,4^5}=0,4^2*2,5^{-4}=(\frac{4}{10})^2*(\frac{25}{10})^{-4}=(\frac{2}{5})^2*(\frac{5}{2})^{-4}=(\frac{2}{5})^2*(\frac{2}{5})^4=(\frac{2}{5})^6}
=0,4^6
d)
{{0,64^5:1,25^{-4}*0,8^{-2}=\frac{((0,8)^2)^5}{1,25^{-4}}*0,8^{-2}=\frac{0,8^{10+(-2)}}{(\frac{125}{100})^{-4}}=(\frac{8}{10})^8:(\frac{5}{4})^{-4}=(\frac{4}{5})^8:(\frac{4}{5})^4=(\frac{4}{5})^4}}
=0,8^4
e)
{\sqrt2 * \sqrt[3]2 : \sqrt8 = \sqrt2 * \frac{\sqrt[3]2}{2\sqrt2} = \frac{2^{\frac{1}{3}}}{2}=2^{\frac{1}{3}-1}=2^{-\frac{2}{3}}=(\frac{1}{2})^{\frac{2}{3}}=\left(\sqrt[3]{\frac{1}{2}}\right)^2=(\frac{1}{\sqrt[3]{2}})^2=(\frac{\sqrt[3]{8}}{\sqrt[3]{2}*\sqrt[3]{4}})^2=(\frac{\sqrt[3]{4}}{2})^2}
f)
{9^{\frac{2}{3}}*\sqrt[4]{27^2}=(3^2)^{\frac{2}{3}}*27^{\frac{2}{4}}=3^{\frac{4}{3}}*(3^3)^{\frac{1}{2}}=3^{\frac{4}{3}+\frac{3}{2}}=3^{\frac{8+9}{6}}=3^{\frac{17}{6}}=(3^{\frac{1}{6}})^{17}=(\sqrt[6]{3})^{17}}