\(A^2=\left(\sqrt{6-x}+\sqrt{x-4}\right)^2\le\left(1+1\right)\left(6-x+x-4\right)=4\)
\(\Rightarrow A\le2\Rightarrow A_{max}=2\) khi \(6-x=x-4\Leftrightarrow x=5\)
\(A\ge\sqrt{6-x+x-4}=\sqrt{2}\)
\(\Rightarrow A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}6-x=0\\x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)