ta có đk \(9-x^2\ge0\Leftrightarrow-3\le x\le3\)
vì \(A=\sqrt{9-x^2}\ge0\)
nên \(0\le A\le3\)
VẬY \(\left\{{}\begin{matrix}A_{min}=0-khi-x=3\\A_{max}=3-khi-x=0\end{matrix}\right.\)
\(A^2=9-x^2;x^2\ge0\Rightarrow A^2\le9-0=9\Rightarrow A\le3\Rightarrow A_{max}=3\)
dâu "=" xayr ra khi: x=0
\(\sqrt{9-x^2}\ge0\forall x\Rightarrow A_{min}=0\)
dâu "=" xayr ra khi: \(x=\pm3\)