Đáp án:
\(\left[ \begin{array}{l}
m > 0\\
m \in \left( { - \infty ; - 2 - \sqrt 3 } \right)
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có hai nghiệm phân biệt cùng âm
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
4{m^2} + 4m + 1 - 3{m^2} > 0\\
\frac{{ - 2m - 1}}{{3m}} < 0\\
\frac{m}{{3m}} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} + 4m + 1 > 0\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 2m - 1 < 0\\
3m > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- 2m - 1 > 0\\
3m < 0
\end{array} \right.
\end{array} \right.\\
\frac{1}{3} > 0\left( {ld} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
m \in \left( { - \infty ; - 2 - \sqrt 3 } \right) \cup \left( { - 2 + \sqrt 3 ; + \infty } \right)\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
m > - \frac{1}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
m < - \frac{1}{2}
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
m \in \left( { - \infty ; - 2 - \sqrt 3 } \right) \cup \left( { - 2 + \sqrt 3 ; + \infty } \right)\\
\left[ \begin{array}{l}
m > 0\\
m < - \frac{1}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m > 0\\
m \in \left( { - \infty ; - 2 - \sqrt 3 } \right)
\end{array} \right.
\end{array}\)
