Đáp án:
\[S = \left( { - 2;1 - \sqrt 2 } \right] \cup \left( {1;1 + \sqrt 2 } \right]\]
Giải thích các bước giải:
ĐKXĐ: \({x^2} + x - 2 \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne - 2
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{{1 - \left| {{x^2} - 2x} \right|}}{{{x^2} + x - 2}} \ge 0\\
TH1:\,\,\,\left\{ \begin{array}{l}
1 - \left| {{x^2} - 2x} \right| \ge 0\\
{x^2} + x - 2 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left| {{x^2} - 2x} \right| \le 1\\
\left( {x - 1} \right)\left( {x + 2} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 1 \le {x^2} - 2x \le 1\\
\left[ \begin{array}{l}
x > 1\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 2x + 1 \ge 0\\
{x^2} - 2x - 1 \le 0\\
\left[ \begin{array}{l}
x > 1\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - \sqrt 2 \le x \le 1 + \sqrt 2 \\
\left[ \begin{array}{l}
x > 1\\
x < - 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow 1 < x \le 1 + \sqrt 2 \\
TH2:\,\,\,\left\{ \begin{array}{l}
1 - \left| {{x^2} - 2x} \right| \le 0\\
{x^2} + x - 2 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left| {{x^2} - 2x} \right| \ge 1\\
\left( {x - 1} \right)\left( {x + 2} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{x^2} - 2x \ge 1\\
{x^2} - 2x \le - 1
\end{array} \right.\\
- 2 < x < 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{x^2} - 2x - 1 \ge 0\\
{x^2} - 2x + 1 \le 0
\end{array} \right.\\
- 2 < x < 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1 + \sqrt 2 \\
x \le 1 - \sqrt 2 \\
x = 1
\end{array} \right.\\
- 2 < x < 1
\end{array} \right.\\
\Leftrightarrow - 2 < x \le 1 - \sqrt 2 \\
\Rightarrow S = \left( { - 2;1 - \sqrt 2 } \right] \cup \left( {1;1 + \sqrt 2 } \right]
\end{array}\)
