Solution: The original equation becomes [1/(x²-10x-29)-1/(x²-10x-69)]+[1/(x²-10x-45)-1/(x²-10x-69)]=0 Let x²-10x-49=a, then x²-10x-29=a+20, x²-10x-69=a-20; x²-10x-57=b, then x²-10x-45=b+12,

Solution: The original equation becomes

[1/(x²-10x-29)-1/(x²-10x-69)]+[1/(x²-10x-45)-1/(x²-10x-69)]=0

Let x²-10x-49=a, then x²-10x-29=a+20, x²-10x-69=a-20; x²-10x-57=b, then x²-10x-45=b+12, x²-10x²-69=b-12

∴a-20=b-12,∴a=b+8

[1/(a+20)-1/(a-20)]+[1/(b+12)-1/(b-12)]=0

[-40/(a²-400)]+[-24/(b²-144)]=0

∴5/(a²-400)+3/(b²-144)=0

∴5b²-720+3a²-1200=0

∴5b²+3a²-1920= 0

∵a=b+8,∴b=a-8

∴5(a-8)²+3a²-1920=0

∴5a²-80a+320+3a²-1920=0

∴8a²-80a-1600=0

∴a²-10a-200=0

(a-20)(a+10)=0

∴a=20 or a= -10

When a=20, x²-10x-49=20, that is, x²-10x-69=0

According to the title, x²-10x-69≠0 (denominator≠0), ∴Give up

When a=-10, x²-10x-49=-10, i.e. x²-10x-39=0, (x+3) (x-13)=0

∴x=-3 or x=13

∴The solution of the original equation is: x1=-3, x2=13

∴The solution of the original equation is: x1=-3, x2=13