Đáp án:
\[\left[ \begin{array}{l}
x < - 2\\
- \frac{1}{2} \le x < 1
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne - 2\\
x \ne 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{{x - 1}}{{x + 2}} \ge \frac{{x + 2}}{{x - 1}}\\
\Leftrightarrow \frac{{x - 1}}{{x + 2}} - \frac{{x + 2}}{{x - 1}} \ge 0\\
\Leftrightarrow \frac{{{{\left( {x - 1} \right)}^2} - {{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} \ge 0\\
\Leftrightarrow \frac{{\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} + 4x + 4} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} \ge 0\\
\Leftrightarrow \frac{{ - 6x - 3}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} \ge 0\\
\Leftrightarrow \frac{{2x + 1}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
x < - 2\\
- \frac{1}{2} \le x < 1
\end{array} \right.
\end{array}\)
