解：$(3)$将方程$2x+\frac {n^2-n}{2x-1}=2n$变形，得$2x-1+\frac {n^2-n}{2x-1} =2n-1$
∴$2x-1+\frac {n(n-1)}{2x-1}=n+n-1$
∴$2x-1=n$或$2x-1=n-1$
∴$x=\frac {n+1}2$或$x=\frac {n}2$
∵$x_{1}＜x_{2}$
∴$x_{1}=\frac {n}2，$$x_{2}=\frac {n+1}2$
∴原式$=\frac {2×\frac {n}2-1}{2×\frac {n+1}2}=\frac {n-1}{n+1}$
