\(A=\frac{3}{2+\sqrt{-x^2+2x+7}}=\frac{3}{2+\sqrt{8-\left(x^2-2x+1\right)}}=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\)
Vì \(\left(x-1\right)^2\le0\Rightarrow8-\left(x-1\right)^2\le8\Rightarrow\sqrt{8-\left(x-1\right)^2}\le\sqrt{8}\)
\(\Rightarrow2+\sqrt{8-\left(x-1\right)^2}\le2+\sqrt{8}\)=>\(A=\frac{3}{2+\sqrt{8-\left(x-1\right)^2}}\ge\frac{3}{2+\sqrt{8}}\)
Dấu "=" xảy ra khi x=1
Vậy minA=\(\frac{3}{2+\sqrt{8}}\) khi x=1
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