Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
Bpt \to \frac{{{x^2} - 4x - 4 - 2{x^2} + 4\left( {m - 1} \right)x - 32}}{{{x^2} - 2\left( {m - 1} \right)x + 16}} \le 0\\
\to \frac{{ - {x^2} + 4\left( {m - 2} \right)x - 36}}{{{x^2} - 2\left( {m - 1} \right) + 16}} \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- {x^2} + 4\left( {m - 2} \right)x - 36 \le 0\\
{x^2} - 2\left( {m - 1} \right) + 16 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- {x^2} + 4\left( {m - 2} \right)x - 36 \ge 0\\
{x^2} - 2\left( {m - 1} \right) + 16 \le 0
\end{array} \right.(l)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 1 \le 0\left( {ld} \right)\\
4\left( {{m^2} - 4m + 4} \right) - 36 \le 0\\
1 \ge 0\left( {ld} \right)\\
{m^2} - 4m + 4 - 16 \le 0
\end{array} \right. \to \left\{ \begin{array}{l}
{m^2} - 4m + 4 \le 9\\
{m^2} - 4m - 12 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 1 \le m \le 5\\
- 2 \le m \le 6
\end{array} \right. \to - 1 \le m \le 5\\
KL: - 1 \le m \le 5
\end{array}\)
