a: \(\text{Δ}=\left(-2m\right)^2-4\cdot1\cdot\left(m^2-m+3\right)\)
\(=4m^2-4m^2+4m-12=4m-12\)
Để phương trình có nghiệm thì Δ>=0
=>4m-12>=0
=>m>=3
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=m^2-m+3\end{matrix}\right.\)
\(Q=x_1^2+x_2^2-4x_1x_2+\left(x_1+x_2\right)\)
\(=\left(x_1+x_2\right)^2-6x_1x_2+\left(x_1+x_2\right)\)
\(=\left(2m\right)^2-6\left(m^2-m+3\right)+2m\)
\(=4m^2-6m^2+6m-18+2m\)
\(=-2m^2+8m-18\)
\(=-2\left(m^2-4m+9\right)\)
\(=-2\left(m^2-4m+4+5\right)\)
\(=-2\left(m-2\right)^2-10< =-10\forall m\)
Dấu '=' xảy ra khi m-2=0
=>m=2