\(A=\left|x\right|\sqrt{\left(9-x^2\right)}\left(đk:-3\le x\le3\right)\\ =\sqrt{x^2\left(9-x^2\right)}\le\dfrac{x^2+9-x^2}{2}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi:
\(x^2=9-x^2\\ \Leftrightarrow2x^2=9\\ \Leftrightarrow x^2=\dfrac{9}{2}\\ \Leftrightarrow x=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
Vậy............