zadanie 1
\frac{16z^2-2}{2,1-6,4*z^3}+z=\frac{16*(\frac{1}{4})^2-2}{2,1-6,4*(\frac{1}{4})^3}+\frac{1}{4}=\frac{16*\frac{1}{16}-2}{2,1-\frac{64}{10}*\frac{1}{64}}+\frac{1}{4}
=\frac{1-2}{2,1-0,1}+\frac{1}{4}=-\frac{1}{2}+\frac{1}{4}=-\frac{2}{4}+\frac{1}{4}=-\frac{1}{4}=-0,25