=>
\(A^2=\left(\frac{\sqrt{\sqrt{10}+3}-\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}-1}}\right)^2\)
\(A^2=\left(\frac{\sqrt{\sqrt{10}+3}-\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}-1}}\right)\left(\frac{\sqrt{\sqrt{10}+3}-\sqrt{\sqrt{10}-3}}{\sqrt{\sqrt{10}-1}}\right)\)
\(A^2=\frac{\sqrt{10}+3+\sqrt{10}-3-2\sqrt{10-9}}{\sqrt{10}-1}\)
\(A^2=\frac{2\left(\sqrt{10}-1\right)}{\sqrt{10}-1}\)
\(A^2=2\)

Thus we have A = √2.
\(B=\sqrt{3+2\sqrt{2}}=\sqrt{\left(2+1\right)-2\sqrt{2\cdot1}}=\sqrt{2}+1\)
Then A – B = √2 – (√2 + 1) = -1.

Therefore, B is the answer.
Answer: B