EDYCJA
\frac{-3}{1+\sqrt2}
\frac{-1}{3+\sqrt3}
\frac{-3}{1+\sqrt2}=\frac{-3(1-\sqrt3)}{(1+\sqrt2)(1-\sqrt2)}=\frac{-3(1-\sqrt3)}{1-2}=\frac{-3(1-\sqrt3)}{-1}=-3(1-\sqrt3)
\frac{-1}{3+\sqrt3}=\frac{-1(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}=\frac{-1(3-\sqrt3}{9-3}=\frac{\sqrt3-3}{6}