Đáp án:
\[ - 1 < m < \frac{5}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) > 0,\,\,\,\,\forall x\\
\Rightarrow \left( {m + 1} \right){x^2} + \left( {2m - 1} \right)x + m - 1 > 0,\,\,\,\,\forall x\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,m = - 1 \Rightarrow f\left( x \right) = - 3x - 2\\
\Rightarrow f\left( x \right) < 0 \Leftrightarrow x > - \frac{2}{3}\,\,\,\,\left( L \right)\\
TH2:\,\,m \ne - 1\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
m + 1 > 0\\
> 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > - 1\\
{\left( {2m - 1} \right)^2} - 4.\left( {m + 1} \right)\left( {m - 1} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > - 1\\
4{m^2} - 4m + 1 - 4{m^2} + 4 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > - 1\\
5 - 4m > 0
\end{array} \right. \Rightarrow - 1 < m < \frac{5}{4}
\end{array}\)
