\(\dfrac{2}{\sqrt{5}-2}+\dfrac{2}{\sqrt{5}+2}=\dfrac{2\sqrt{5}+4+2\sqrt{5}-4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{4\sqrt{5}}{21}\\ \dfrac{\sqrt{3}\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}=\dfrac{\sqrt{3}\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}=\sqrt{3}-\dfrac{1}{2-\sqrt{3}}\\ =\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}=\left(2\sqrt{3}-4\right)\left(2-\sqrt{3}\right)\\ =8\sqrt{3}-14\)
1) \(=\dfrac{2\left(\sqrt{5}+2\right)+2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{2\sqrt{5}+4+2\sqrt{5}-4}{5-4}=4\sqrt{5}\)
2) \(=\dfrac{\sqrt{3}.\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{2+\sqrt{3}}{4-3}=\sqrt{3}-2-\sqrt{3}=-2\)
