Đáp án: m>0 hoặc m<-4
Giải thích các bước giải:
$\begin{array}{l}
m\left( {m + 2} \right){x^2} + 2mx + 2 > 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m\left( {m + 2} \right) > 0\\
\Delta ' < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
{m^2} - 2.m\left( {m + 2} \right) < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
- {m^2} - 4m < 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 0\\
m < - 2
\end{array} \right.\\
\left[ \begin{array}{l}
m > 0\\
m < - 4
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
m > 0\\
m < - 4
\end{array} \right.
\end{array}$
