Ta có: \(\sqrt{10}\ge x\ge-\sqrt{10}\)
\(\left(x+3\right)\sqrt{10-x^2}< x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\left[\sqrt{10-x^2}-\left(x-4\right)\right]< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3< 0\\\sqrt{10-x^2}-\left(x-4\right)>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3>0\\\sqrt{10-x^2}-\left(x-4\right)< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -3\\\sqrt{10-x^2}>x-4\left(Luôn-đúng\forall x< -3\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>-3\\-\sqrt{10}< x< -3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt{10}< x< -3\)
Vậy ..........
