\(\text{Δ}=\left[2\left(m+1\right)\right]^2-4\left(m^2-m\right)\)
\(=4m^2+8m+4-4m^2+4m=12m+4\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>12m+4>0
=>12m>-4
=>\(m>-\dfrac{1}{3}\)
Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2;x_1x_2=m^2-m\)
\(\left(x_1-1\right)^2+\left(x_2-1\right)^2=2\)
=>\(\left(x_1^2+x_2^2\right)-2\left(x_1+x_2\right)=0\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-2\left(x_1+x_2\right)=0\)
=>\(\left(2m+2\right)^2-2\left(2m+2\right)-2\left(m^2-m\right)=0\)
=>\(4m^2+8m+4-4m-4-2m^2+2m=0\)
=>\(2m^2+6m=0\)
=>m(m+3)=0
=>\(\left[{}\begin{matrix}m=0\left(nhận\right)\\m=-3\left(loại\right)\end{matrix}\right.\)
