Đáp án:
C2:
m>0
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
\left( {m + 1} \right){x^2} + mx + m < 0\forall x\\
\to \left\{ \begin{array}{l}
m + 1 < 0\\
{m^2} - 4m\left( {m + 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < - 1\\
{m^2} - 4{m^2} - 4m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < - 1\\
- 3{m^2} - 4m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < - 1\\
\left[ \begin{array}{l}
m > 0\\
m < - \dfrac{4}{3}
\end{array} \right.
\end{array} \right.\\
\to m < - \dfrac{4}{3}\\
\to C\\
C2:\\
\left( {m + 1} \right){x^2} + mx + m < 0\\
DK:\left\{ \begin{array}{l}
m + 1 > 0\\
{m^2} - 4m\left( {m + 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
{m^2} - 4{m^2} - 4m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
- 3{m^2} - 4m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
\left[ \begin{array}{l}
m > 0\\
m < - \dfrac{4}{3}
\end{array} \right.
\end{array} \right.\\
\to m > 0
\end{array}\)
