Zadanie 1
a)
\frac{(\frac{1}{4})^{-1}*2^3*\sqrt{16}}{32^{-\frac{2}{3}}*0,25}=\frac{4*2^3*4}{(2^5)^{-\frac{2}{3}}*\frac{1}{4}}=\frac{16*2^3}{2^{-\frac{10}{3}}*4^{-1}}=\frac{2^4*2^3}{2^{-\frac{10}{3}}*2^{-2}}=
=\frac{2^7}{2^{-\frac{10}{3}-\frac{6}{3}}}=\frac{2^7}{2^{-\frac{16}{3}}}=2^{\frac{21}{3}-(-\frac{16}{3})}=2^{-\frac{37}{3}}
b)
\sqrt[4]{4\sqrt[3]4} = 4^{\frac{1}{4}}*(4^{\frac{1}{3}})^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}*(2^2)^{\frac{1}{12}}=2^{\frac{1}{2}}*2^{\frac{1}{6}}=
=2^{\frac{3}{6}+\frac{1}{6}}=2^{\frac{4}{6}}=2^{\frac{2}{3}}