\(\left|x-1\right|=x^2+2x-4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=x^2+2x-4\left(1\right)\\x-1=-\left(x^2+2x-4\right)\left(2\right)\end{matrix}\right.\)
Giải (1) : \(x-1=x^2+2x-4\)
\(\Leftrightarrow x-1-x^2-2x+4=0\)
\(\Leftrightarrow-x^2-x+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
Giải (2) : \(x-1=-\left(x^2+2x-4\right)\)
\(\Leftrightarrow x-1=-x^2-2x+4\)
\(\Leftrightarrow x-1+x^2+2x-4=0\)
\(\Leftrightarrow x^2+3x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{29}}{2}\\x=\dfrac{-3-\sqrt{29}}{2}\end{matrix}\right.\)
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