Pt có 2 nghiệm
\(\to \Delta=(-2m)^2-4.1.(m^2-m+3)=4m^2-4m^2+4m-12=4m-12\ge 0\\\leftrightarrow 4m\ge 12\\\leftrightarrow m\ge 3\)
Theo Viét
\(\begin{cases}x_1+x_2=2m\\x_1x_2=m^2-m+3\end{cases}\)
\( (2x_2-1)x_1+(2x_1-1)x_2\\=2x_1x_2-x_1+2x_1x_2-x_2\\=4x_1x_2-(x_1+x_2)\\=4.(m^2-m+3)-2m\\=4m^2-4m+12-2m\\=4m^2-6m+12\\=4\bigg(m^2-\dfrac{6}{4}m+3\bigg)\\=4\bigg(m^2-2.\dfrac{3}{2}.m+\dfrac{9}{4}+\dfrac{3}{4}\bigg)\\=4\bigg(m-\dfrac{3}{2}\bigg)^2+3\ge 3\\\to\A_{\min}=3\\\to m-\dfrac{3}{2}=0\\\leftrightarrow m=\dfrac{3}{2}\)
Vậy \(A_{\min}=3\)