Để pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}m\ne0\\\Delta=\left(2m-1\right)^2-4m\left(m-2\right)>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m>-\frac{1}{4}\\m\ne0\end{matrix}\right.\)
Theo Viet ta có: \(x_1+x_2=\frac{2m-1}{m}=2003\)
\(\Rightarrow2m-1=2003m\Rightarrow m=-\frac{1}{2001}\) (t/m)