ĐK $x\ne 2$
Ta có $(\sqrt 2 +1)^x \ge \bigg ( \dfrac{1}{\sqrt 2 +1)} \bigg )^{\frac{x}{x-2}}$
$\Leftrightarrow \bigg ( \sqrt 2 + 1 \bigg )^{\frac{x^2 -x}{x-2}} \ge 1 =(\sqrt 2 +1)^0$
$\Leftrightarrow \frac{x^2 -x}{x-2} \ge 0$
$\Leftrightarrow \left [ \begin{matrix} x >2 \\ \\ 0 \le x \le 1 \end{matrix} \right.$
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