Đáp án:
\[m \le - \frac{2}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
m{x^2} - 2\left( {m - 1} \right)x + 3m + 2 < 0,\,\,\,\,\,\,\forall x \in \left( { - \infty ;0} \right)\\
\Leftrightarrow m{x^2} - 2mx + 2x + 3m + 2 < 0,\,\,\,\,\,\,\forall x \in \left( { - \infty ;0} \right)\\
\Leftrightarrow m\left( {{x^2} - 2x + 3} \right) + \left( {2x + 2} \right) < 0,\,\,\,\,\,\,\forall x \in \left( { - \infty ;0} \right)\\
\Leftrightarrow m\left( {{x^2} - 2x + 3} \right) < - \left( {2x + 2} \right),\,\,\,\,\,\,\forall x \in \left( { - \infty ;0} \right)\\
\Leftrightarrow m < \frac{{ - \left( {2x + 2} \right)}}{{{x^2} - 2x + 3}},\,\,\,\,\,\,\forall x \in \left( { - \infty ;0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} - 2x + 3 = {{\left( {x - 1} \right)}^2} + 2 > 0} \right)\\
\Leftrightarrow m \le \mathop {\min }\limits_{\left( { - \infty ;0} \right)} f\left( x \right) = \frac{{ - \left( {2x + 2} \right)}}{{{x^2} - 2x + 3}} = f\left( 0 \right) = - \frac{2}{3}\\
\Rightarrow m \le - \frac{2}{3}
\end{array}\)
