If $$x=6+2\sqrt{6}$$, then what is the value of $$\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$$ ?
We need to calculate $$\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$$
This equals $$\frac{x-1 + 1}{\sqrt{x-1}} = \frac{x}{\sqrt{x-1}}$$
$$x-1 = 5+2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$$
Therefore, $$\sqrt{x-1} = \sqrt{3} + \sqrt{2}$$
Hence, the required expression becomes $$\frac{6+2\sqrt{6}}{\sqrt{2}+\sqrt{3}}$$
This equals $$2*\frac{3+\sqrt{6}}{\sqrt{2}+\sqrt{3}}=2\sqrt{3}$$
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