Đáp án:
\[m \ge \frac{1}{3}\]
Giải thích các bước giải:
Phương trình \(f\left( x \right) = a\,{x^2} + bx + c \ge 0,\,\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta \le 0
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
f\left( x \right) \ge 0,\,\,\,\forall x \in R\\
\Leftrightarrow m{x^2} - 2\left( {m - 1} \right)x + 4m \ge 0,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta ' \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{\left( {m - 1} \right)^2} - m.4m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - 2m + 1 - 4{m^2} \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
- 3{m^2} - 2m + 1 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
3{m^2} + 2m - 1 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\left( {m + 1} \right)\left( {3m - 1} \right) \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m \ge \frac{1}{3}\\
m \le - 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow m \ge \frac{1}{3}
\end{array}\)
Vậy \(m \ge \frac{1}{3}\)
