\(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\left(-2m+1\right)\)
\(=4\left(m-1\right)^2-4\left(-2m+1\right)\)
\(=4\left(m^2-2m+1+2m-1\right)=4m^2\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(4m^2>0\)
=>\(m^2\ne0\)
=>\(m\ne0\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=-2m+1\end{matrix}\right.\)
Vì \(a+b+c=1+2m-2-2m+1=0\)
nên phương trình luôn có hai nghiệm là:
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=-2m+1\end{matrix}\right.\)
\(2x_1+3x_2=7\)
=>\(\left[{}\begin{matrix}2\cdot1+3\left(-2m+1\right)=7\\2\left(-2m+1\right)+3\cdot1=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3\left(2m+1\right)=5\\2\left(-2m+1\right)=4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2m+1=-\dfrac{5}{3}\\-2m+1=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2m=-\dfrac{8}{3}\\-2m=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=\dfrac{4}{3}\left(nhận\right)\\m=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
