a: Ta có: \(A=-2\sqrt{3}\left(3-\sqrt{3}\right)+\left(3\sqrt{3}+1\right)^2\)
\(=-6\sqrt{3}+12+28+6\sqrt{3}\)
=40
b: Ta có: \(B=\dfrac{8+2\sqrt{2}}{3-\sqrt{2}}-\dfrac{2+3\sqrt{2}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{2}-1}\)
\(=4+2\sqrt{2}-\sqrt{2}-3-2-\sqrt{2}\)
=-1