\(A=\sqrt{6}-\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(\Leftrightarrow\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}-A\)
\(\Leftrightarrow2+\sqrt{3}+2-\sqrt{3}+2\cdot\sqrt{\left(\sqrt{2+\sqrt{3}}\right)\cdot\left(\sqrt{2-\sqrt{3}}\right)}=A^2+6-2A\sqrt{6}\)
\(\Leftrightarrow4+2\cdot1=A^2+6-2A\sqrt{6}\)
\(\Leftrightarrow6=A^2+6+2A\sqrt{6}\)
\(\Leftrightarrow A^2+2A\sqrt{6}=0\Rightarrow\left\{{}\begin{matrix}A^2=0\\2A\sqrt{6}=0\end{matrix}\right.\Leftrightarrow A=0\)
Vậy A = 0.
\(A^2=6+2+\sqrt{3}+2-\sqrt{3}-2\sqrt{6}.\sqrt{2+\sqrt{3}}-2\sqrt{6}.\sqrt{2-\sqrt{3}}+2\\ =12-\sqrt{48+24\sqrt{3}}-\sqrt{48-24\sqrt{3}}\\ =12-\sqrt{\left(6+2\sqrt{3}\right)^2}-\sqrt{\left(6-2\sqrt{3}\right)^2}\\ =12-6-2\sqrt{3}-6+2\sqrt{3}=0\Rightarrow A=0\)
