\(\Leftrightarrow\left(x-2\right)\left(\sqrt{x^2+2}-x-2\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-2\ge0\\\sqrt{x^2+2}\ge x+2\end{matrix}\right.\) ko tồn tại x thỏa mãn
TH2: \(\left\{{}\begin{matrix}x-2\le0\\\sqrt{x^2+2}\le x+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le2\\x^2+2\le x^2+4x+4\end{matrix}\right.\)
\(\Rightarrow-\frac{1}{2}\le x\le2\)