+ \(A=\sqrt{-x^2+2x+4}=\sqrt{-\left(x^2-2x+1\right)+5}=\sqrt{-\left(x-1\right)^2+5}\le\sqrt{5}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
+ \(A=\sqrt{-x^2+2x+4}\ge0\forall x\)
Dấu "=" \(\Leftrightarrow-x^2+2x+4=0\Leftrightarrow\left(x-1\right)^2=5\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{matrix}\right.\)