ĐK: x khác 0
Ta có pt tương đương \(x^2+2+\dfrac{1}{x^2}-\left(x+\dfrac{1}{x}\right)-12=0\)
<=>\(\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-12=0\)
<=>\(\left[{}\begin{matrix}x+\dfrac{1}{x}=-3\\x+\dfrac{1}{x}=4\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x^2+3x+1=0\left(1\right)\\x^2-4x+1=0\left(2\right)\end{matrix}\right.\)
Ta thấy pt (1)vô nghiệm
Pt(2) <=>\(\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
KL \(x=2+\sqrt{3}\), \(x=2-\sqrt{3}\)