Tìm m để hs luôn có cực trị : y= mx mũ 3 – x mũ 2 +2m mũ2 x +2m -3

\[\begin{array}{l}
y = m{x^3} – {x^2} + 2{m^2}x + 2m – 3\\
\Rightarrow y’ = 3m{x^2} – 2x + 2{m^2}\\
\Rightarrow y’ = 0\\
\Leftrightarrow 3m{x^2} – 2x + 2{m^2} = 0\,\,\left( * \right)\\
TH1:\,\,Hs\,\,luon\,\,co\,\,\,cuc\,\,tri \Leftrightarrow \left( * \right)\,\,\,co\,\,2\,\,nghiem\,\,phan\,\,biet\\
\Leftrightarrow \left\{ \begin{array}{l}
a \ne 0\\
\Delta ‘ > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
1 – 6{m^3} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^3} < \frac{1}{6} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 0\\ m < \frac{1}{{\sqrt[3]{6}}} \end{array} \right..\\ TH2:\,\,m = 0\\ \Rightarrow y = - {x^2} - 3\,\,\,la\,\,\,ham\,\,so\,\,bac\,\,\,hai\,\,co\,\,1\,\,cuc\,\,\,tri.\\ Vay\,\,m < \frac{1}{{\sqrt[3]{6}}}\,\,\,thoa\,\,\,man\,\,bai\,\,toan.\, \end{array}\]