Odpowiedź:
[tex]\dfrac{\sqrt{3}}{\sqrt{6}}=\dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\\\dfrac{4}{\sqrt{3}-1}=\dfrac{4}{\sqrt{3}-1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{3}+1}=\dfrac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{4(\sqrt{3}+1)}{(\sqrt{3})^2-1^2}=\dfrac{4(\sqrt{3}+1)}{3-1}=\\\\\\=\dfrac{\not4^2(\sqrt{3}+1)}{\not2_{1}}=2(\sqrt{3}+1)=2\sqrt{3}+2[/tex]
[tex]\dfrac{1}{\sqrt{2}+1}=\dfrac{1}{\sqrt{2}+1}\cdot\dfrac{\sqrt{2}-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^2-1^2}=\dfrac{\sqrt{2}-1}{2-1}=\\\\\\=\dfrac{\sqrt{2}-1}{1}=\sqrt{2}-1[/tex]
