The value of $\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\frac{1}{\sqrt{5-2\sqrt{6}}}$ is

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$$\begin{gathered} \sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\frac{1}{\sqrt{5-2\sqrt{6}}} \\ First let \\ I=\frac{1}{\sqrt{5-2\sqrt{6}}} \\ =\frac{1}{\sqrt{5-2\sqrt{6}}}\times \frac{\sqrt{5-2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}} \left(Multiply by the conjugate \right) \\ =\frac{\sqrt{5-2\sqrt{6}}}{5-2\sqrt{6}} \\ =\frac{\sqrt{5-2\sqrt{6}}}{5-2\sqrt{6}}\times \frac{5+2\sqrt{6}}{5+2\sqrt{6}} \left(Multiply by the conjugate \right) \\ =\frac{\left(\sqrt{5-2\sqrt{6}}\right)\left(5+2\sqrt{6}\right)}{{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{25-24} \\ =\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}} \\ So , \\ \sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\frac{1}{\sqrt{5-2\sqrt{6}}} \\ =\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}} \\ =\sqrt{2\times 3+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}} \\ =\boxed{\sqrt{\mathbf{2}\left(\mathbf{3}\mathbf{+}\sqrt{\mathbf{3}}\mathbf{+}\sqrt{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{6}}\right)}\mathbf{-}\left(\mathbf{5}\mathbf{+}\mathbf{2}\sqrt{\mathbf{6}}\right)\sqrt{\mathbf{5}\mathbf{-}\mathbf{2}\sqrt{\mathbf{6}}}} \end{gathered}$$

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