\(x^2+\left|x\right|-6=0\)
\(\Rightarrow\left|x\right|^2+\left|x\right|-6=0\)
Đặt: \(\left|x\right|=t\ge0\) ta có: \(pt\Leftrightarrow t^2+t-6=0\)
\(\Rightarrow t^2+t+\dfrac{1}{4}-\dfrac{25}{4}=0\)
\(\Rightarrow\left(t+\dfrac{1}{2}\right)^2-\dfrac{25}{4}=0\)
\(\Rightarrow\left(t+\dfrac{1}{2}+\dfrac{5}{2}\right)\left(t+\dfrac{1}{2}-\dfrac{5}{2}\right)=0\)
\(\Rightarrow\left(t+3\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-3\left(loại\right)\\t=2\end{matrix}\right.\)
Vậy \(\left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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