Bài 2:
\(\text{Δ}=\left[-\left(2m+1\right)\right]^2-4\cdot1\cdot2m\)
\(=\left(2m+1\right)^2-8m\)
\(=4m^2+4m+1-8m\)
\(=4m^2-4m+1=\left(2m-1\right)^2\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>\(\left(2m-1\right)^2>0\)
=>\(2m-1\ne0\)
=>\(2m\ne1\)
=>\(m\ne\dfrac{1}{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-\left(2m+1\right)\right]}{1}=2m+1\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m}{1}=2m\end{matrix}\right.\)
\(x_1^2+x_2^2-x_1x_2=1\)
=>\(\left(x_1+x_2\right)^2-3x_1x_2=1\)
=>\(\left(2m+1\right)^2-3\cdot2m-1=0\)
=>\(4m^2+4m+1-6m-1=0\)
=>\(4m^2-2m=0\)
=>2m(2m-1)=0
=>\(\left[{}\begin{matrix}m=0\left(nhận\right)\\m=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
