\(A=\sqrt{11+2\sqrt{24}}=2\sqrt{2}+\sqrt{3}\)

\(B=\dfrac{4}{2+\sqrt{4-2\sqrt{3}}}=\dfrac{4}{2+\sqrt{3}-1}=\dfrac{4}{\sqrt{3}+1}=2\left(\sqrt{3}-1\right)=2\sqrt{3}-2\)

\(A=\sqrt{11+\sqrt{96}}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=2\sqrt{2}+\sqrt{3}\)

\(B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=1+\sqrt{2}+\sqrt{3}\)

Xét : \(A-B=2\sqrt{2}+\sqrt{3}-\left(1+\sqrt{2}+\sqrt{3}\right)=\sqrt{2}-1>0\)

\(\Rightarrow A>B\)

\(B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}.\)\(=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}=1+\sqrt{2}+\sqrt{3}\)

\(A=\sqrt{11+\sqrt{96}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}\)\(=2\sqrt{2}+\sqrt{3}>1+\sqrt{2}+\sqrt{3}=B\)

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 \(1< 2\Leftrightarrow\sqrt{1}< \sqrt{2}\Leftrightarrow1< \sqrt{2}\Leftrightarrow\sqrt{3}+\sqrt{2}+1< \sqrt{3}+\sqrt{2}+\sqrt{2}\Leftrightarrow A>B\)

\(B=\sqrt{11+2\sqrt{24}}-4\sqrt{2}-\sqrt{3}\)

\(=2\sqrt{2}+\sqrt{3}-4\sqrt{2}-\sqrt{3}\)

\(=-2\sqrt{2}\)

Ta có: \(\sqrt{14-4\sqrt{6}}-\frac{\sqrt{5}+1}{\sqrt{2}}+\sqrt{11+\sqrt{96}}+\sqrt{3-\sqrt{5}}\)

\(=\sqrt{12-2\cdot2\sqrt{3}\cdot\sqrt{2}+2}-\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{22+2\sqrt{96}}}{\sqrt{2}}+\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(=\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}+\frac{-\sqrt{5}-1+\sqrt{16+2\cdot4\cdot\sqrt{6}+6}+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}{\sqrt{2}}\)

\(=\left|2\sqrt{3}-\sqrt{2}\right|+\frac{-\sqrt{5}-1+\sqrt{\left(4+\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)

\(=2\sqrt{3}-\sqrt{2}+\frac{-\sqrt{5}-1+\left|4+\sqrt{6}\right|+\left|\sqrt{5}-1\right|}{\sqrt{2}}\)(Vì \(2\sqrt{3}>\sqrt{2}\))

\(=2\sqrt{3}-\sqrt{2}+\frac{-\sqrt{5}-1+4+\sqrt{6}+\sqrt{5}-1}{\sqrt{2}}\)(Vì \(\left\{{}\begin{matrix}4>\sqrt{6}>0\\\sqrt{5}>1\end{matrix}\right.\))

\(=2\sqrt{3}-\sqrt{2}+\frac{2+\sqrt{6}}{\sqrt{2}}\)

\(=2\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}\)

\(=3\sqrt{3}\)

\(A=\sqrt{11+\sqrt{96}}>B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)

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