Đáp án:
c. \(m \in \left[ {1; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
m < 0\\
{m^2} - 2m + 1 - 4m\left( {m - 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
- 3{m^2} + 2m + 1 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 0\\
m \in \left( { - \infty ; - \dfrac{1}{3}} \right) \cup \left( {1; + \infty } \right)
\end{array} \right.\\
\to m \in \left( { - \infty ; - \dfrac{1}{3}} \right)\\
c.DK:\left( {m + 1} \right){x^2} - 2\left( {m - 1} \right)x + 3m - 3 \ge 0\forall x \in R\\
\to \left\{ \begin{array}{l}
m + 1 > 0\\
{m^2} - 2m + 1 - 3\left( {m + 1} \right)\left( {m - 1} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
{m^2} - 2m + 1 - 3\left( {{m^2} - 1} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
- 2{m^2} - 2m + 4 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - 1\\
m \in \left( { - \infty ; - 2} \right] \cup \left[ {1; + \infty } \right)
\end{array} \right.\\
\to m \in \left[ {1; + \infty } \right)
\end{array}\)
