b: Thay x=-2 vào (1), ta được:
\(\left(-2\right)^2-2\left(m-1\right)\cdot\left(-2\right)+m-3=0\)
=>\(4+4\left(m-1\right)+m-3=0\)
=>4m-4+m+1=0
=>5m-3=0
=>m=3/5
c: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\left(m-3\right)\)
\(=4m^2-8m+4-4m+12\)
\(=4m^2-12m+16=\left(2m-3\right)^2+7>=7>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=m-3\end{matrix}\right.\)
\(A=x_1^2+x_2^2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(2m-2\right)^2-2\left(m-3\right)\)
\(=4m^2-8m+4-2m+6\)
\(=4m^2-10m+10\)
\(=\left(2m\right)^2-2\cdot2m\cdot2,5+6,25+3,75\)
\(=\left(2m-2,5\right)^2+3,75>=3,75\forall m\)
Dấu '=' xảy ra khi 2m-2,5=0
=>m=1,25
