\(A^2=\left(2+\sqrt{3}\right)^2=7+\sqrt{48}\)
\(B^2=\left(\sqrt{2}+3\right)^2=11+\sqrt{72}\)
\(\hept{\begin{cases}7< 11\\\sqrt{48}< \sqrt{72}\end{cases}\Leftrightarrow}7+\sqrt{48}< 11+\sqrt{72}\)
\(\Rightarrow A< B\)
Ta có:\(2+\sqrt{3}< 2+\sqrt{4}=4=\sqrt{1}+3< \sqrt{2}+3\)
\(\Rightarrow2+\sqrt{3}< \sqrt{2}+3\)