\(A=\sqrt{x-2}+\sqrt{6-x}\le\sqrt{\left(1+1\right)\left(x-2+6-x\right)}=2\sqrt{2}\)
\(\Rightarrow A_{max}=2\sqrt{2}\) khi \(x-2=6-x\Leftrightarrow x=4\)
Lại có
\(A^2=x-2+2\sqrt{\left(x-2\right)\left(6-x\right)}+6-x=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)
\(\Rightarrow A\ge2\) \(\Rightarrow A_{min}=2\) khi \(x=2\) hoặc \(x=6\)