Đáp án:
$\begin{array}{l}
f\left( x \right) = \left( {m + 1} \right){x^2} + \left( {2m - 1} \right)x + m - 1\\
f\left( x \right) \ge 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m + 1 > 0\\
\Delta < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > - 1\\
{\left( {2m - 1} \right)^2} - 4\left( {m + 1} \right).\left( {m - 1} \right) < 0\left( * \right)
\end{array} \right.\\
\left( * \right) \Rightarrow 4{m^2} - 4m + 1 - 4\left( {{m^2} - 1} \right) < 0\\
\Rightarrow 4{m^2} - 4m + 1 - 4{m^2} + 4 < 0\\
\Rightarrow 4m > 5\\
\Rightarrow m > \frac{5}{4}\\
Vậy\,m > \frac{5}{4}
\end{array}$
