\(x^4+4x^2-x^2-4=0\Rightarrow\left(x^4-x^2\right)+\left(4x^2-4\right)=0\)
\(\Rightarrow x^2\left(x^2-1\right)+4\left(x^2-1\right)=0\Rightarrow\left(x^2+4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+4=0\\x-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}vn\\x=1\\x=-1\end{matrix}\right.\)
Vậy pt có 2 nghiệm: \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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