Ta có
\(A=\sqrt{11+\sqrt{96}}=\sqrt{11+\sqrt{16.6}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=\left|2\sqrt{2}+\sqrt{3}\right|=2\sqrt{2}+\sqrt{3}=\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(B=\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{1+2\sqrt{2}+2-3}=\dfrac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=1+\sqrt{2}+\sqrt{3}\)
Ta có \(2>1\Leftrightarrow\sqrt{2}>\sqrt{1}\Leftrightarrow\sqrt{2}>1\Leftrightarrow\sqrt{2}+\sqrt{2}+\sqrt{3}>1+\sqrt{2}+\sqrt{3}\)\(\Leftrightarrow\sqrt{11+\sqrt{96}}>\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
Vậy A>B