czy to tak wygląda?
\frac{(x-y\sqrt{3})^2-(x+y\sqrt{3})^2}{2xy^2-6x^2y}=\frac{x^2-2xy\sqrt{3}+(y\sqrt{3})^2-[x^2+2xy\sqrt{3}+(y\sqrt{3})^2}{2xy(y-3x)}=\frac{x^2-2xy\sqrt{3}+3y^2-x^2-2xy\sqrt{3}-3y^2}{2xy(y-3x)}=\frac{-4xy\sqrt{3}}{2xy(y-3x)}=\frac{-2\sqrt{3}}{y-3x}
\frac{-2\sqrt{3}}{-2-3*1}=\frac{-2\sqrt{3}}{-5}=\frac{-2}{-5}\sqrt{3}=0,4\sqrt{3}