Đáp án:
\[ - 1 \le x \le 1\]
Giải thích các bước giải:
ĐKXĐ: \(x \ne 0\)
Ta có:
\(\begin{array}{l}
{x^2} \le \left| {1 - \frac{2}{{{x^2}}}} \right|\\
\Leftrightarrow {x^2} \le \left| {\frac{{{x^2} - 2}}{{{x^2}}}} \right|\\
\Leftrightarrow {x^2} \le \frac{{\left| {{x^2} - 2} \right|}}{{{x^2}}}\\
\Leftrightarrow {x^4} \le \left| {{x^2} - 2} \right|\,\,\,\,\,\,\,\,\,\,\left( 1 \right)
\end{array}\)
\(\begin{array}{l}
TH1:\,\,\,\,{x^2} - 2 < 0 \Leftrightarrow - \sqrt 2 < x < \sqrt 2 \\
\left( 1 \right) \Leftrightarrow {x^4} \le 2 - {x^2}\\
\Leftrightarrow {x^4} + {x^2} - 2 \le 0\\
\Leftrightarrow \left( {{x^2} - 1} \right)\left( {{x^2} + 2} \right) \le 0\\
\Leftrightarrow {x^2} \le 1\\
\Leftrightarrow - 1 \le x \le 1\,\,\,\,\left( {t/m} \right)\\
TH2:\,\,{x^2} - 2 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 2 \\
x \le - \sqrt 2
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow {x^4} \le {x^2} - 2\\
\Leftrightarrow {x^4} - {x^2} + 2 \le 0\,\,\,\,\,\left( {vn} \right)
\end{array}\)
Vậy \( - 1 \le x \le 1\)
