Answer:
Step-by-step explanation:
[tex]\frac{1}{2+\sqrt{3}}=\frac{1*(2-\sqrt{3})}{(2+\sqrt{3})*(2-\sqrt{3})}\\\\=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}=\frac{2-\sqrt{3}}{4-3}\\\\=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}\\\\\frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2*(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})*(\sqrt{5} +\sqrt{3})}\\\\=\frac{2*(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}=\frac{2*(\sqrt{5}+\sqrt{3})}{5-3}\\\\=\frac{2*(\sqrt{5}+\sqrt{3})}{2}=\sqrt{5}+\sqrt{3}\\\\[/tex]
[tex]\frac{1}{2-\sqrt{5}}=\frac{1*(2+\sqrt{5})}{(2-\sqrt{5})*(2+\sqrt{5})}\\\\=\frac{2+\sqrt{5}}{(2)^{2}-(\sqrt{5})^{2}}\\\\=\frac{2+\sqrt{5}}{4-5}=\frac{2+\sqrt{5}}{-1}\\\\= -2-\sqrt{5}\\\\\\\\\frac{1}{2+\sqrt{3}}+\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{1}{2-\sqrt{5}}=2-\sqrt{3}+\sqrt{5}+\sqrt{3}-2-\sqrt{5}=0[/tex]
