解方程1/(x^2-2x-2)+25/(x^2-2x+2)=24/(x^2-2x+1)

由题意不妨令t=x²-2x+1=(x-1)²,那么：
原方程可化为1/(t-3) + 25/(t+1)=24/t
上式左边通分得：
(t+1+25t-75)/[(t-3)(t+1)]=24/t
(26t-74)/[(t-3)(t+1)]=24/t
(13t-37)/[(t-3)(t+1)]=12/t
t(13t-37)=12(t-3)(t+1)
13t²-37t=12t²-24t-36
t²-13t+36=
(t-4)(t-9)=0
解得：t=4或t=9
那么：(x-1)²=4或(x-1)²=9
所以：x-1=2或x-1=-2或x-1=3或x-1=-3
解得：x=3或x=-1或x=4或x=-2
经检验可知x=3或x=-1或x=4或x=-2均是原方程的解.