Đáp án:
\[m \le \frac{{ - 3 - 2\sqrt 3 }}{3}\]
Giải thích các bước giải:
\(\begin{array}{l}
\left( {m + 1} \right){x^2} + \left( {m - 1} \right)x + 2m \le 0,\,\,\,\,\forall x \in R\,\,\,\,\,\left( * \right)\\
TH1:\,\,\,m + 1 = 0 \Leftrightarrow m = - 1\\
\left( * \right) \Leftrightarrow - 2x - 2 \le 0,\,\,\,\,\forall x \in R\\
\Leftrightarrow x \ge - 1,\,\,\,\,\forall x \in R\,\,\,\,\,\,\,\,\,\left( L \right)\\
TH2:\,\,\,m \ne - 1\\
\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
m + 1 < 0\\
Δ \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
{\left( {m - 1} \right)^2} - 4.\left( {m + 1} \right).m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
{m^2} - 2m + 1 - 4{m^2} - 4m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
- 3{m^2} - 6m + 1 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
3{m^2} + 6m - 1 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < - 1\\
\left[ \begin{array}{l}
m \ge \frac{{ - 3 + 2\sqrt 3 }}{3}\\
m \le \frac{{ - 3 - 2\sqrt 3 }}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow m \le \frac{{ - 3 - 2\sqrt 3 }}{3}
\end{array}\)
