Đáp án:

c. \(m \in \left( { - 15; - 4} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left\{ \begin{array}{l}
3m + 1 < 0\\
9{m^2} + 6m + 1 - 4.\left( {3m + 1} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - \left( {12m + 4} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - 12{m^2} - 52m - 16 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
 - 3{m^2} - 46m - 15 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
 - \left( {3m + 1} \right)\left( {m + 15} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
\left( {3m + 1} \right)\left( {m + 15} \right) > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - \dfrac{1}{3}\\
m \in \left( { - \infty ; - 15} \right) \cup \left( { - \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
 \to m \in \left( { - \infty ; - 15} \right)\\
b.\left\{ \begin{array}{l}
3m + 1 > 0\\
9{m^2} + 6m + 1 - 4.\left( {3m + 1} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - \left( {12m + 4} \right)\left( {m + 4} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
9{m^2} + 6m + 1 - 12{m^2} - 52m - 16 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
 - 3{m^2} - 46m - 15 < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
 - \left( {3m + 1} \right)\left( {m + 15} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
\left( {3m + 1} \right)\left( {m + 15} \right) > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - \dfrac{1}{3}\\
m \in \left( { - \infty ; - 15} \right) \cup \left( { - \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
 \to m \in \left( { - \dfrac{1}{3}; + \infty } \right)\\
c.Ycbt \to \left\{ \begin{array}{l}
9{m^2} + 6m + 1 - 4.\left( {3m + 1} \right)\left( {m + 4} \right) > 0\\
\dfrac{{m + 4}}{{3m + 1}} > 0\\
\dfrac{{3m + 1}}{{3m + 1}} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - \left( {3m + 1} \right)\left( {m + 15} \right) > 0\\
m \in \left( { - \infty ; - 4} \right) \cup \left( { - \dfrac{1}{3}; + \infty } \right)\\
1 > 0\left( {ld} \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left( {3m + 1} \right)\left( {m + 15} \right) < 0\\
m \in \left( { - \infty ; - 4} \right) \cup \left( { - \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \in \left( { - 15; - \dfrac{1}{3}} \right)\\
m \in \left( { - \infty ; - 4} \right) \cup \left( { - \dfrac{1}{3}; + \infty } \right)
\end{array} \right.\\
 \to m \in \left( { - 15; - 4} \right)
\end{array}\)