Observed \sqrt{2}, \sqrt{3} and \sqrt{6} appears in both terms of the expression.
Expand the square of \sqrt{2}+\sqrt{3} for a study.
(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}
Then the first term could be simplified as
\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6} }
=\sqrt{(\sqrt{2}+\sqrt{3} )^2+2(\sqrt{2}+\sqrt{3})+1 }
=\sqrt{(\sqrt{2}+\sqrt{3} +1)^2}
=\sqrt{2}+\sqrt{3} +1
Multiply numerator and denominator with the conjugate of the radicant. The second term is simplified by using of difference of squares formula.
\dfrac{1}{\sqrt{5-2\sqrt{6}} }
=\dfrac{1}{\sqrt{5-2\sqrt{6} }}\cdot \dfrac{\sqrt{5+2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}
=\sqrt{5+2\sqrt{6}}
=(\sqrt{\sqrt{2}+\sqrt{3} )^2}
=\sqrt{2}+\sqrt{3}
Then, subtracting the second term from the first term gives
\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6} }-\dfrac{1}{5-2\sqrt{6} }
=\sqrt{2}+\sqrt{3} +1 - (\sqrt{2}+\sqrt{3} )
=1
Therefore, none of the choices are correct. D is the right choice.