\(\Delta=\left(-2m\right)^2-4\cdot1\cdot\left(-4m-5\right)\)
\(=4m^2+16m+20\)
\(=4\left(m^2+4m+5\right)=4\left(m^2+4m+4+1\right)\)
\(=4\left(m+2\right)^2+4>=4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=-4m-5\end{matrix}\right.\)
\(x_1^2-2\left(m-1\right)x_1+2x_2-4m=5+2x_1x_2\)
=>\(x_1^2-2m\cdot x_1+2x_1+2x_2-4m=5+2x_1x_2\)
=>\(x_1^2-x_1\left(x_1+x_2\right)+2\left(x_1+x_2\right)-4m-5-2x_1x_2=0\)
=>\(-x_1x_2+2\left(x_1+x_2\right)-4m-5-2x_1x_2=0\)
=>\(2\left(x_1+x_2\right)-3x_1x_2-4m-5=0\)
=>\(2\cdot2m-3\cdot\left(-4m-5\right)-4m-5=0\)
=>\(4m+12m+15-4m-5=0\)
=>12m+10=0
=>12m=-10
=>\(m=-\dfrac{5}{6}\)
