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

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

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: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)

=10

Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)

\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)

\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)

\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)

a)So sánh vs 5/2

b)So sánh vs 40/9

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