Bài 2:
=>|2x^2-3x-1|=1-2x
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x^2-3x-1\right)^2=\left(1-2x\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x^2-3x-1-2x+1\right)\left(2x^2-3x-1+2x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x^2-5x\right)\left(2x^2-x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{1-\sqrt{17}}{4}\right\}\)
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