Đáp án:
$S=\{\pm2\sqrt2\}$
Giải thích các bước giải:
$\begin{array}{l}\quad x^2 +3x + 1 = (x+3)\sqrt{x^2+1}\\ \Leftrightarrow x(x+3) - (x+3)\sqrt{x^2 +1} +1 =0\\ \Leftrightarrow (x+3)(x - \sqrt{x^2 + 1}) + 1 =0\\ \Leftrightarrow (x+3)\cdot\dfrac{(x-\sqrt{x^2+1})(x + \sqrt{x^2 +1)}}{x + \sqrt{x^2+1}} + 1 =0 \\ \Leftrightarrow (x+3)\cdot\dfrac{-1}{x+\sqrt{x^2+1}} +1 =0\\ \Leftrightarrow \dfrac{x+3}{x + \sqrt{x^2+1}} = 1\\ \Leftrightarrow x + 3 = x + \sqrt{x^2+1}\\ \Leftrightarrow \sqrt{x^2 +1}= 3\\\Rightarrow x^2 + 1 = 9\\ \Leftrightarrow x^2 = 8\\ \Leftrightarrow x = \pm2\sqrt2\\ Vậy\,\,S=\{\pm2\sqrt2\} \end{array}$
