Đáp án:
2) \(m \in \left( {1;3} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)m{x^2} - 2mx + 3m + 4 \ge 0\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - m\left( {3m + 4} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - 3{m^2} - 4m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
- 2{m^2} - 4m \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
- 2m\left( {m + 2} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 0\\
m \in \left( { - \infty ; - 2} \right] \cup \left[ {0; + \infty } \right)
\end{array} \right.\\
\to m > 0\\
2)f(x) = {x^2} - 2(m + 1)x + 6m - 2 > 0\forall x \in R\\
\to {m^2} + 2m + 1 - 6m + 2 < 0\\
\to {m^2} - 4m + 3 < 0\\
\to \left( {m - 1} \right)\left( {m - 3} \right) < 0\\
\to m \in \left( {1;3} \right)
\end{array}\)
