Đáp án:
\[\left[ \begin{array}{l}
\frac{{11}}{4} \le x \le 3\\
x \le - 1
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: D=R
Ta thấy: \({x^2} - 5x + 7 > 0,\,\,\,\,\forall x \in D\) do đó, ta có:
\(\begin{array}{l}
\frac{1}{{13}} \le \frac{{{x^2} - 2x - 2}}{{{x^2} - 5x + 7}} \le 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2} - 2x - 2}}{{{x^2} - 5x + 7}} \ge \frac{1}{{13}}\\
\frac{{{x^2} - 2x - 2}}{{{x^2} - 5x + 7}} \le 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
13\left( {{x^2} - 2x - 2} \right) \ge {x^2} - 5x + 7\\
{x^2} - 2x - 2 \le {x^2} - 5x + 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
12{x^2} - 21x - 33 \ge 0\\
3x \le 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \frac{{11}}{4}\\
x \le - 1
\end{array} \right.\\
x \le 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\frac{{11}}{4} \le x \le 3\\
x \le - 1
\end{array} \right.
\end{array}\)
