解方程(x+1)⼀(x-1)-(x-1)⼀(x+1)=1

[（x+1）²-（x-1）²]／（x-1）（x+1）=1
[（x+1）²-（x-1）²]=（x-1）（x+1）
x²+2x+1-x²+2x-1=x²-1
-x²+4x+1=0
x²-4x-1=0
x²-4x+4=5
(x-2)²=5

x1= 2+√5； x2=2-√5
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(x+1)/(x-1)-(x-1)/(x+1)=1

乘以（x-1)(x+1)得
（x+1)²-(x-1)²=x²-1
x²+2x+1-(x²-2x+1)=x²-1
x²-4x-1=0
x²-4x+4=5
(x-2)²=5
x-2=±√5
x1=2+√5, x2=2-√5

(x+1)/(x-1)-(x-1)/(x+1)=1
:同*(x-1)(x+1)
[(x+1)^2-(x-1)^2]/[(x-1)(x+1)]=1
(x+1-x+1)(x+1+x-1)/(x^2-1)=1
4x=x^2-1
x^2-4x-1=0
x1=(4+根(16+4))/2=2+根5 x2=2-根5

(x+1)/(x-1)-(x-1)/(x+1)=1
(x+1)^2-(x-1)^2=(x+1)(x-1)
4xy=x^2-1
x^2-4xy-1=0
x^2-4xy+4=5
(x-2)^2=5
x-2=±根号5
x=2±根号5

(x+1)²/(x²-1)-(x-1)²/(x²-1)=1
[(x+1)²-(x-1)²]/(x²-1)=1
4x/(x²-1)=1
4x=x²-1
x²-4x+4=5
(x-2)²=(√5)²
x-2=±√5
x=2±√5

等号两边都乘以(x+1)(x-1)可以得到：
(x+1)²-(x-1)²=(x+1)(x-1)
得到：x²+2x+1-x²+2x-1=x²-1
合并同类项可得：
x²-4x-1=0

即：x²-4x+4=5

(x-2)²=5
得到最后的结果为：
x1= 2+√5 或 x2=2-√5