[tex]Zadanie\\\\\frac{2}{\sqrt{2}+\sqrt{3}-1}=\frac{2}{(\sqrt{2}+\sqrt{3})-1} \cdot \frac{(\sqrt{2}+\sqrt{3})+1}{(\sqrt{2}+\sqrt{3})+1}= \frac{2\sqrt{2}+2\sqrt{3}+2}{(\sqrt{2}+\sqrt{3})^{2}- 1^{2} }=\frac{2\sqrt{2}+2\sqrt{3}+2}{2+2\sqrt{6}+3-1}=\\\\= \frac{2\sqrt{2}+2\sqrt{3}+2}{2\sqrt{6}+4}=\frac{2\sqrt{2}+2\sqrt{3}+2}{2(\sqrt{6}+2) }\cdot\frac{\sqrt{6}-2}{\sqrt{6}-2}=\frac{2\sqrt{12}+2\sqrt{18}+2\sqrt{6}-4\sqrt{2}- 4\sqrt{3}-4}{2[(\sqrt{6})^{2}-2^{2}]}=[/tex]
[tex]=\frac{4\sqrt{3}+6\sqrt{2}+2\sqrt{6}-4\sqrt{2}-4\sqrt{3}-4}{2(6-4)}=\frac{2\sqrt{2}+ 2\sqrt{6}-4}{4}=\frac{\sqrt{2}+\sqrt{6}-2}{2}[/tex]
