Đáp án:
7d) \(m \le - \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
6c)f\left( x \right) < 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
- 1 < 0\\
4\left( {{m^2} + 2m + 1} \right) + 1 - {m^2} < 0
\end{array} \right.\\
\to 3{m^2} + 8m + 5 < 0\\
\to \left( {m + 1} \right)\left( {3m + 5} \right) < 0\\
\to - \dfrac{5}{3} < m < - 1\\
7d)DK:\left\{ \begin{array}{l}
m < 0\\
{m^2} - 2m + 1 - 4m\left( {m - 1} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
{m^2} - 2m + 1 - 4{m^2} + 4m \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
- 3{m^2} + 2m + 1 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
\left[ \begin{array}{l}
m \ge 1\\
m \le - \dfrac{1}{3}
\end{array} \right.
\end{array} \right.\\
KL:m \le - \dfrac{1}{3}
\end{array}\)
