Đáp án:
b) \( - 3 < m < \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Ycbt \Leftrightarrow \left( {{m^2} + 4m - 5} \right){x^2} - 2\left( {m - 1} \right)x + 2 \ge 0\forall x\\
\to \left\{ \begin{array}{l}
{m^2} + 4m - 5 > 0\\
{m^2} - 2m + 1 - 2\left( {{m^2} + 4m - 5} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 5
\end{array} \right.\\
{m^2} - 2m + 1 - 2{m^2} - 8m + 10 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 5
\end{array} \right.\\
- {m^2} - 10m + 11 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 1\\
m < - 5
\end{array} \right.\\
\left[ \begin{array}{l}
m \ge 1\\
m \le - 11
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
m > 1\\
m \le - 11
\end{array} \right.\\
b)Ycbt \Leftrightarrow \left( {2 - 3m} \right){x^2} + 2mx + m - 1 > 0\\
\to \left\{ \begin{array}{l}
2 - 3m > 0\\
{m^2} + 2m - 3 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{2}{3} > m\\
- 3 < m < 1
\end{array} \right.\\
\to - 3 < m < \dfrac{2}{3}\\
c)Ycbt \Leftrightarrow m{x^2} + \left( {m + 2} \right)x + 2 \ge 0\\
\to \left\{ \begin{array}{l}
m > 0\\
{m^2} + 4m + 4 - 4m.2 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 0\\
{m^2} + 4m + 4 - 8m \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 0\\
{m^2} - 4m + 4 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 0\\
{\left( {m - 2} \right)^2} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 0\\
m - 2 = 0
\end{array} \right.\\
\to m = 2\left( {TM} \right)
\end{array}\)
