Đáp án:
$S = \left\{\pm \sqrt{\dfrac{\sqrt{8045} -1}{2}}\right\}$
Giải thích các bước giải:
$\begin{array}{l}\quad x^4 + \sqrt{x^2 + 2012} = 2012\\ Đặt\,\,t = \sqrt{x^2 + 2012}\qquad (t >0)\\ \to t^2 =x^2 + 2012\\ \to (t^2 -2012)^2 = x^4\\ \text{Phương trình trở thành:}\\ (t^2 - 2012)^2 + t = 2012\\ \to t^4 - 4024t^2 + 2012^2 + t - 2012 =0\\ \to t^4 -t^3 -2011t^2 + t^3 -t^2 - 2011t -2012t^2 + 2012t + 2012.2011 = 0\\ \to t^2(t^2 - t - 2011) + t(t^2 - t - 2011) - 2012(t^2 - t - 2011) = 0\\ \to (t^2 - t - 2011)(t^2 + t -2012)= 0\\ \to \left[\begin{array}{l}t = \dfrac{1 -\sqrt{8045}}{2}\quad (loại)\\t = \dfrac{1 +\sqrt{8045}}{2}\quad (nhận)\\t = \dfrac{-1 -\sqrt{8049}}{2}\quad (loại)\\t = \dfrac{-1 +\sqrt{8049}}{2}\quad (nhận)\end{array}\right.\\ +) \quad t = \dfrac{1 +\sqrt{8045}}{2}\\ \to \sqrt{x^2 + 2012} = \dfrac{1 +\sqrt{8045}}{2}\\ \to x^2 = \dfrac{\sqrt{8045} -1}{2}\\ \to x = \pm \sqrt{\dfrac{\sqrt{8045} -1}{2}}\\ +) \quad t = \dfrac{-1 +\sqrt{8049}}{2}\\ \to \sqrt{x^2 + 2012} = \dfrac{-1 +\sqrt{8049}}{2}\\ \to x^2 = \dfrac{1 - \sqrt{8049}}{2} \quad \text{(vô nghiệm)}\\ Vậy\,\,S = \left\{\pm \sqrt{\dfrac{\sqrt{8045} -1}{2}}\right\} \end{array}$
