\(\Delta=\left(2n-1\right)^2-4n\left(n-1\right)\)
\(=4n^2-4n+1-4n^2+4n=1\)
Phương trình có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=\frac{2n-1-1}{2}=n-1\\x_2=\frac{2n-1+1}{2}=n\end{matrix}\right.\)
\(x_1^2-2x_2+3=\left(n-1\right)^2-2n+3=n^2-4n+4=\left(n-2\right)^2\ge0\) (đpcm)