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Ta có:
$x^2+\sqrt{2x^2+4x+3}\geq 6-2x$
$\Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 6-2x-x^2<0\\ 2x^2+4x+3\ge0 \end{array} \right.\\\left\{ \begin{array}{l} 6-2x-x^2\ge0\\ 2x^2+4x+3\ge(6-2x-x^2)^2 \end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x>\sqrt7-1\\ x<-\sqrt7-1\\ \left\{ \begin{array}{l} -\sqrt7-1\le x\le\sqrt7-1\\(x-1)(x+3)(x^2+2x-11)\le0 \end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l} x\ge 1\\ x\le-3 \end{array} \right.$
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