\(\text{Δ}=\left(2m\right)^2-4\cdot1\cdot\left(m^2-m+1\right)\)
\(=4m^2-4m^2+4m-4=4m-4\)
Để phương trình có hai nghiệm thì Δ>=0
=>4m-4>=0
=>m>=1
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2m\\x_1x_2=\dfrac{c}{a}=m^2-m+1\end{matrix}\right.\)
\(A=x_1^2-2m\cdot x_2+2x_1+2x_2+2026\)
\(=x_1^2+x_2\left(x_1+x_2\right)+2\left(x_1+x_2\right)+2026\)
\(=\left(x_1^2+x_2^2\right)+x_1x_2+2\left(x_1+x_2\right)+2026\)
\(=\left(x_1+x_2\right)^2-x_1x_2+2\left(x_1+x_2\right)+2026\)
\(=\left(-2m\right)^2-\left(m^2-m+1\right)+2\cdot\left(-2m\right)+2026\)
\(=4m^2-m^2+m-1-4m+2026\)
\(=3m^2-3m+2025\)
\(=3\left(m^2-m+675\right)\)
\(=3\left(m^2-m+\dfrac{1}{4}+674,75\right)\)
\(=3\left(m-\dfrac{1}{2}\right)^2+\dfrac{8097}{4}>=\dfrac{8097}{4}\forall m\)
Dấu '=' xảy ra khi \(m-\dfrac{1}{2}=0\)
=>\(m=\dfrac{1}{2}\)
