Đặt x2 = t > 0 ta được
\(2t+1=\dfrac{1}{t}-4\Leftrightarrow2t^2+5t-1=0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\\t=\dfrac{-5-\sqrt{33}}{4}\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy pt có 2 nghiệm
\(2x^2+1=\dfrac{1}{x^2}-4\left(1\right)\)
Đặt \(x^2=t\left(t\ge0\right)\)
Khi đó phương trình \(\left(1\right)\) trở thành \(2t+1=\dfrac{1}{t}-4\)
\(\Leftrightarrow2t^2+5t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\left(\text{nhận}\right)\\t=\dfrac{-5-\sqrt{33}}{4}\left(\text{loại}\right)\end{matrix}\right.\)
\(\Rightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-\sqrt{-5+\sqrt{33}}}{2};\dfrac{\sqrt{-5+\sqrt{33}}}{2}\right\}\)
