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Remind the following properties.
\(\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\)
\(\sqrt{a+b-2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\left(a>b\right)\)
\(\left(\frac{\sqrt{\sqrt{10}+3}+\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}+1}}\right)^2\)
\(\frac{\left(\sqrt{\sqrt{10}+3}+\sqrt{\sqrt{10}-3}\right)^2}{\left(\sqrt{\sqrt{10}+1}\right)^2}\)
\(\frac{\left(\sqrt{10}+3\right)+\left(\sqrt{10}-3\right)+2\sqrt{\left(\sqrt{10}+3\right)\left(\sqrt{10}-3\right)}}{\sqrt{10}+1}\)
\(\frac{\left(\sqrt{10}+3\right)+\left(\sqrt{10}-3\right)+2\sqrt{10-3\sqrt{10}+3\sqrt{10}-9}}{\sqrt{10}+1}\)
\(\frac{2\sqrt{10}+2}{\sqrt{10}+1}\)
\(\frac{2\left(\sqrt{10}+1\right)}{\sqrt{10}+1}\)
=2

Thus, we have \(\frac{\sqrt{\sqrt{10}+3}+\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}+1}}=\sqrt{2}\)

Since we have \(\sqrt{7-2\sqrt{10}}=\sqrt{\left(5+2\right)-2\sqrt{5\cdot2}}=\sqrt{5}-\sqrt{2}\) , we have

\(\frac{\sqrt{\sqrt{10}+3}+\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}+1}}+\sqrt{7-2\sqrt{10}}=\sqrt{2}+\left(\sqrt{5}-\sqrt{2}\right)=\sqrt{5}\)

Therefore, D is the answer.
Answer: D