Given

\sqrt{2-\sqrt{3} } +\dfrac{2}{\sqrt{3} +1+\sqrt{2} }

The expression is the sum of two terms with square root.

Let's start from the first term

\sqrt{2-\sqrt{3} } = \sqrt{\dfrac{4-2\sqrt{3} }{2} } = \sqrt{\dfrac{(\sqrt{3-1} )^2}{2} }=\dfrac{\sqrt{3}-1 }{\sqrt{2} } =\dfrac{\sqrt{6}-\sqrt{2} }{2}

Simplify the second term

\dfrac{2}{\sqrt{3} +1+\sqrt{2} }

=\dfrac{2}{\sqrt{3}+\sqrt{2}+1}\cdot \dfrac{\sqrt{3}+\sqrt{2}-1}{\sqrt{3}+\sqrt{2}-1}

=\dfrac{2(\sqrt{3}+\sqrt{2}-1)}{(\sqrt{3}+\sqrt{2})^2-1}

=\dfrac{2(\sqrt{3}+\sqrt{2}-1)}{4+2\sqrt{6} }

=\dfrac{\sqrt{3}+\sqrt{2}-1}{\sqrt{6} +2 }\cdot \dfrac{\sqrt{6} -2 }{\sqrt{6}-2 }

=\dfrac{ 3\sqrt{2}+2\sqrt{3}-\sqrt{6} -2\sqrt{3}-2\sqrt{2}+2 }{2}

=\dfrac{\sqrt{2} -\sqrt{6} +2 }{2}

Addition of the two terms yields

\dfrac{\sqrt{6}-\sqrt{2} }{2} +\dfrac{\sqrt{6}-\sqrt{2} +2 }{2} = 1

So the value of \sqrt{2-\sqrt{3} } +\dfrac{2}{\sqrt{3} +1+\sqrt{2} } is equal to 1