Ta có \(A=\sqrt{11+\sqrt{96}}=\sqrt{11+\sqrt{16.6}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=2\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}+\sqrt{2}\)Ta lại có \(B=\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\dfrac{2\sqrt{2}\left[1-\left(\sqrt{2}-\sqrt{3}\right)\right]}{\left(1+\sqrt{2}-\sqrt{3}\right)\left[1-\left(\sqrt{2}-\sqrt{3}\right)\right]}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1^2-\left(\sqrt{2}-\sqrt{3}\right)^2}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(2-2\sqrt{6}+3\right)}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{1-5+2\sqrt{6}}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{2\sqrt{6}-4}=\dfrac{2\sqrt{2}\left(1-\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{1-\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)\(=\dfrac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{6}-2+3+\sqrt{6}}{3-2}=\sqrt{3}+\sqrt{2}+1\)Vì \(1< 2\Leftrightarrow\sqrt{1}< \sqrt{2}\Leftrightarrow1< \sqrt{2}\Leftrightarrow\sqrt{3}+\sqrt{2}+1< \sqrt{3}+\sqrt{2}+\sqrt{2}\Leftrightarrow A>B\)
