Đáp án:
\[m > 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) > 0,\,\,\,\forall x \in R\\
\Leftrightarrow \left( {m - 1} \right){x^2} + \left( {4m - 3} \right)x + 5m - 3,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 > 0\\
Δ\le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
{\left( {4m - 3} \right)^2} - 4.\left( {m - 1} \right).\left( {5m - 3} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
16{m^2} - 24m + 9 - 4.\left( {5{m^2} - 8m + 3} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
- 4{m^2} - 3 \le 0
\end{array} \right. \Leftrightarrow m > 1
\end{array}\)
