Lời giải:
Ta có:
\(2x-x^2+7=8-(x^2-2x+1)=8-(x-1)^2\leq 8\)
\(\Rightarrow \sqrt{2x-x^2+7}\leq 2\sqrt{2}\)
\(\Rightarrow 2+\sqrt{2x-x^2+7}\leq 2+2\sqrt{2}\)
\(\Rightarrow A=\frac{3}{2+\sqrt{2x-x^2+7}}\geq \frac{3}{2+2\sqrt{2}}\)
Vậy $A_{\min}=\frac{3}{2+2\sqrt{2}}$ tại $(x-1)^2=0$ hay $x=1$