\(A=\frac{x^2-3}{x+2}\)
\(\Leftrightarrow A\left(x+2\right)=x^2-3\)
\(\Leftrightarrow Ax+2A=x^2-3\)
\(\Leftrightarrow x^2-xA-2A-3=0\)
\(\Delta=A^2-4\cdot\left(-2A-3\right)\)
\(=A^2+8A+12\ge0\)
\(\Leftrightarrow\left(A+4\right)^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}A+4\ge2\\A+4\le-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}A\ge-2\\A\le-6\end{matrix}\right.\)
Vậy \(A_{min}=-2\Leftrightarrow x=-1\). \(A_{max}=-6\Leftrightarrow x=-3\)