\(\left(\left|x^2-1\right|\right)=\left(x-1\right)^2\\ x^2-1=x^2-2x+1\\ x^2-1-x^2+2x-1=0\\ 2x-2=0\\ 2x=2\\x=2:2\\ x=1\)
Sai thì thôi á
`|x^2-1|=x-1`
`<=>`\(\left[{}\begin{matrix}x-1=x^2-1\\x-1=1-x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1-x^2+1=0\\x-1-1+x^2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x-x^2=0\\x^2+x-1=0\end{matrix}\right.\)
`@ TH1 : x-x^2=0`
`<=> x(1-x)=0 ` \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
`@ TH2 : x^2 +x-1 =0`
`=> Delta = 1^2 -4 .1.(-1) = 5>0`
`=>` PT của 2 nghiệm riêng biệt`
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-1+\sqrt{5}}{2}\\x_2=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
