Đáp án:
\(m \in \left( { - \infty ; - \frac{1}{3}} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
f\left( x \right) \ge 0\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - m > 0\\
{m^2} - 4m + 4 - \left( {2 - m} \right)\left( { - 4m + 1} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 2\\
{m^2} - 4m + 4 + 8m - 2 - 4{m^2} + m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < 2\\
- 3{m^2} + 5m + 2 \le 0\left( 1 \right)
\end{array} \right.\\
Xét: - 3{m^2} + 5m + 2 = 0\\
\to \left[ \begin{array}{l}
m = 2\\
m = - \frac{1}{3}
\end{array} \right.
\end{array}\)
BXD:
x -∞ \( - \frac{1}{3}\) 2 +∞
(1) - 0 + 0 -
\(\begin{array}{l}
Bpt:\left( 1 \right) \to m \in \left( { - \infty ; - \frac{1}{3}} \right] \cup \left[ {2; + \infty } \right)\\
KL:m \in \left( { - \infty ; - \frac{1}{3}} \right]
\end{array}\)
