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根据方程特点，考虑引入中间量
令x-7=y,则 1/(y+3)+1/(y-3)=1/(y-1)+1/(y+1)
( (y-3)+(y+3))/(y-3)(y+3)=((y+1)+(y-1))/(y-1)(y+1)
2y/(y-3)(y+3)=2y/(y-1)(y+1)
2y/(y^2-9)=2y/(y^2-1)
y=0
即x-7=0
x=7
令x-7=y,则 1/(y+3)+1/(y-3)=1/(y-1)+1/(y+1)
( (y-3)+(y+3))/(y-3)(y+3)=((y+1)+(y-1))/(y-1)(y+1)
2y/(y-3)(y+3)=2y/(y-1)(y+1)
2y/(y^2-9)=2y/(y^2-1)
y=0
即x-7=0
x=7
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（x-8)(x-6)(x-10)+(x-4)(x-8)(x-6)=(x-4)(x-10)(x-6)+(x-4)(x-10）（x-8)
(x-8)(x-6)(2x-14)=(x-4)(x-10)(2x-14)
(2x-14)(x^2-14x+48-x^2+14x-40)=0
2(x-7)*8=0
x=7
(x-8)(x-6)(2x-14)=(x-4)(x-10)(2x-14)
(2x-14)(x^2-14x+48-x^2+14x-40)=0
2(x-7)*8=0
x=7
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