a: \(A=2\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)+2\sqrt{3}\)
\(=4+2\sqrt{3}\)
\(B=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\sqrt{x}-1\)
b: Để A<B thì \(\sqrt{x}-1>4+2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}>5+2\sqrt{3}\)
hay \(x>37+20\sqrt{3}\)

