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