a)
[tex]\left(\sqrt{6}-\sqrt{5}\right)^{-1}=\dfrac{1}{\sqrt{6}-\sqrt{5}}=\dfrac{1}{\sqrt{6}-\sqrt{5}}\cdot\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\dfrac{\sqrt{6}+\sqrt{5}}{6-5}=[/tex]
[tex]=\dfrac{\sqrt{6}+\sqrt{5}}{1}=\sqrt{6}+\sqrt{5}[/tex]
b)
[tex]\left(2+\sqrt{3}\right)^{-1}=\dfrac{1}{2+\sqrt{3}}=\dfrac{1}{2+\sqrt{3}}\cdot\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=\dfrac{2-\sqrt{3}}{1}=2-\sqrt{3}[/tex]
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a)
[tex]\left(\sqrt{6}-\sqrt{5}\right)^{-1}=\dfrac{1}{\sqrt{6}-\sqrt{5}}=\dfrac{1}{\sqrt{6}-\sqrt{5}}\cdot\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\dfrac{\sqrt{6}+\sqrt{5}}{6-5}=[/tex]
[tex]=\dfrac{\sqrt{6}+\sqrt{5}}{1}=\sqrt{6}+\sqrt{5}[/tex]
b)
[tex]\left(2+\sqrt{3}\right)^{-1}=\dfrac{1}{2+\sqrt{3}}=\dfrac{1}{2+\sqrt{3}}\cdot\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=\dfrac{2-\sqrt{3}}{1}=2-\sqrt{3}[/tex]