\(\Leftrightarrow\left[{}\begin{matrix}3\left(m+6\right)x^2-3\left(m+3\right)x+2m-3>3\\3\left(m+6\right)x^2-3\left(m+3\right)x+2m-3< -3\end{matrix}\right.\) \(\forall x\)
\(\Leftrightarrow\left[{}\begin{matrix}3\left(m+6\right)x^2-3\left(m+3\right)x+2m-6>0\left(1\right)\\3\left(m+6\right)x^2-3\left(m+3\right)x+2m< 0\left(2\right)\end{matrix}\right.\)
\(m=-6\) ko thỏa mãn
TH1: xét (1)
\(\Leftrightarrow\left\{{}\begin{matrix}m+6>0\\9\left(m+3\right)^2-12\left(m+6\right)\left(2m-6\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m>-6\\5m^2+6m-171>0\end{matrix}\right.\) \(\Rightarrow m>\frac{-3+12\sqrt{6}}{5}\)
TH2: xét (2)
\(\Leftrightarrow\left\{{}\begin{matrix}m< -6\\9\left(m+3\right)^2-24m\left(m+6\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< -6\\5m^2+30m-27>0\end{matrix}\right.\) \(\Rightarrow m< \frac{-15-6\sqrt{10}}{5}\)
Lấy hợp 2 nghiệm (xấu quá)
