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ĐK: $x^2+3x+1\geq 0 \Leftrightarrow x\in R\backslash \left( \frac{-3-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}\right)$
Đặt $\sqrt{x^2+3x+1}=t\geq 0$ thì PT tương đương với:
$\sqrt{t^2+5}+t=\sqrt{3t^2+13}$
$\Leftrightarrow \frac{t^2-4}{\sqrt{t^2+5}+3}+t-2=\frac{3(t^2-4)}{\sqrt{3t^2+13}+5}$
$\Leftrightarrow (t-2)\left[ (t+2)\left( \frac{1}{\sqrt{t^2+5}+3}-\frac{1}{\sqrt{3t^2+13}+5}\right) +1\right] =0$
$\Leftrightarrow t=2$
$\Leftrightarrow x^2+3x-3=0$
$\Leftrightarrow x=\frac{-3\pm \sqrt{21}}{2}$
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