\(x^4-x^2-2=0\)
\(\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow\left(x^4-2x^2\right)+\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2\right)+\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right).\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[x^2-\left(\sqrt{2}\right)^2\right].\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right).\left(x+\sqrt{2}\right).\left(x^2+1\right)=0\)
Vì \(x^2\ge0\) \(\forall x.\)
\(\Rightarrow x^2+1>0\) \(\forall x.\)
\(\Rightarrow x^2+1\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\x+\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+\sqrt{2}\\x=0-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\sqrt{2};-\sqrt{2}\right\}.\)
Chúc bạn học tốt!
Đặt: \(x^2=t\) ta có:
\(t^2-t-2=0\left(t>0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)
\(Với:t=2\Rightarrow x=\sqrt{2}\)
Vậy .................
