\[\begin{array}{l}
y = m{x^3} + m{x^2} + x - 1\\
\Rightarrow y' = 3m{x^2} + 2mx + 1\\
\Rightarrow y' = 0 \Leftrightarrow 3m{x^2} + 2mx + 1 = 0\,\,\,\left( * \right)\\
a)\,\,\,Hs\,\,\,DB\,\,\,tren\,\,\left( {2; + \infty } \right).\\
TH1:\,\,\,m = 0\\
\Rightarrow y' = 1 > 0 \Rightarrow hs\,\,\,DB\,\,tren\,\,\,R.\\
\Rightarrow m = 0\,\,tm\,\,bai\,\,toan.\\
TH2:\,\,m \ne 0\\
+ )\,\,hs\,\,\,Db\,\,\,tren\,\,\,R\\
\Leftrightarrow y' \ge 0\,\,\forall x \in R\\
\Leftrightarrow \Delta ' \le 0\,\,\forall x \in R\\
\Leftrightarrow {m^2} - 3m \le 0\\
\Leftrightarrow 0 \le m \le 3.\\
\Rightarrow 0 < m \le 3\,\,\,thoa\,\,man\,\,\,bai\,\,toan.\\
+ )\,\,\,pt\,\,y' = 0\,\,\,co\,\,2\,\,\,nghiem\,\,pb\\
\Leftrightarrow \Delta ' > 0 \Leftrightarrow {m^2} - 2m > 0 \Leftrightarrow \left[ \begin{array}{l}
m > 3\\
m < 0
\end{array} \right..\\
Pt\,\,co\,\,2\,\,\,nghiem\,\,pb\,\,\,{x_1},\,\,{x_2}\,\,\,ta\,\,co:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \frac{2}{3}\\
{x_1}{x_2} = \frac{1}{{3m}}
\end{array} \right.\\
Ta\,\,\,co\,\,bang\,\,xet\,\,dau:\\
TH\,\,m > 3:\\
\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_2}\,\,\,\,\,\,\,\,\,\,\,\, + \\
\Rightarrow hs\,\,\,DB\,\,tren\,\,\,\,\left( {2; + \infty } \right) \Leftrightarrow {x_1} < {x_2} \le 2\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} \le 4\\
\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} \le 4\\
{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- \frac{2}{3} \le 4\,\,\left( {luon\,\,dung} \right)\\
\frac{1}{{3m}} - 2.\frac{{ - 2}}{3} + 4 \ge 0
\end{array} \right. \Leftrightarrow \frac{1}{{3m}} \ge - \frac{{16}}{3} \Leftrightarrow m \ge - \frac{1}{{16}}\\
\Rightarrow m > 3\,\,thoa\,\,\,man\,\,bai\,\,toan.\\
TH\,\,m < 0\\
\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,{x_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_2}\,\,\,\,\,\,\,\,\,\,\,\, - \\
\Rightarrow hs\,\,k\,\,the\,\,\,DB\,\,tren\,\,\left( {2;\,\, + \infty } \right).
\end{array}\]
Câu b em làm tương tự nhé em.
