Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:
\(x^2+x-4+\dfrac{1}{x}+\dfrac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)+x+\dfrac{1}{x}-6=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+x+\dfrac{1}{x}-6=0\)
Đặt \(x+\dfrac{1}{x}=t\)
\(\Rightarrow t^2+t-6=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x+1=0\\x^2+3x+1=0\end{matrix}\right.\) (bấm máy)
