$解：∵2x+\frac{n^2+n-2}{2x+1}=2n$
$∴2x+1+\frac{n^2+n-2}{2x+1}=2n+1$
$2x+1+\frac{(n+2)(n-1)}{2x+1}=(n+2)+(n-1)$
$∴2x+1=n+2或2x+1=n-1$
$∴x=\frac{n+1}{2}或x=\frac{n-2}{2}$
$∵x_1＜x_2$
$∴x_1=\frac{n-2}{2}，x_2=\frac{n+1}{2}$
$∴原式= \frac{2×\frac{n-2}{2}}{2×\frac{n+1}{2}-3}=\frac{n-2}{n-2}=1$
