\(\text{Δ}=\left[2\left(m-1\right)\right]^2-4\cdot1\cdot\left(-2m\right)\)
\(=\left(2m-2\right)^2+8m=4m^2+4>=4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=-2m\end{matrix}\right.\)
\(\left|x_1+1\right|=\left|x_2+1\right|\)
=>\(\left[{}\begin{matrix}x_1+1=x_2+1\\x_1+1=-x_2-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x_1=x_2\\x_1+x_2=-2\end{matrix}\right.\)
TH1: \(x_1=x_2\)
mà \(x_1+x_2=-2\left(m-1\right)\)
nên \(x_1=x_2=-\left(m-1\right)\)
\(x_1x_2=-2m\)
=>\(-2m=\left(m-1\right)^2\)
=>\(m^2-2m+1=-2m\)
=>\(m^2+1=0\)(vô lý)
TH2: \(x_1+x_2=-2\)
=>-2(m-1)=-2
=>m-1=1
=>m=2
