解分式方程x^2-3x+2/x-3+x^2-9x+2/x-9=x^2-5x+2/x-5+x^2-7x+2/x-7

(x^2-3x+2)/(x-3)+(x^2-9x+2)/(x-9)=(x^2-5x+2)/(x-5)+(x^2-7x+2)/(x-7)
x+2/(x-3)+x+2/(x-9)=x+2/(x-5)+x+2/(x-7)
2/(x-3)+2/(x-9)=2/(x-5)+2/(x-7)
1/(x-3)+1/(x-9)=1/(x-5)+1/(x-7)
1/(x-9)-1/(x-7)=1/(x-5)-1/(x-3)
[(x-7)-(x-9)]/(x-9)(x-7)=[(x-3)-(x-5)]/(x-5)(x-3)
2/(x-9)(x-7)= 2/(x-5)(x-3)
(x-5)(x-3)=(x-9)(x-7)
x²-8x+15=x²-16x+63
8x=48
x=6
检验：x=6是方程的根1/(x-3)+1/(x-9)=1/(x-5)+1/(x-7)[(x-9)+(x-3)]/(x-3)(x-9)=[(x-7)+(x-5)]/(x-5)(x-7)2x-12/(x-3)(x-9)=2x-12/(x-5)(x-7)(x-3)(x-9)=(x-5)(x-7)x^2-12x+27=x^2-12x+35以后就解不下去，哪里错了？两边不可以约去2x-12是不是分母无解，分子2x-12只能等于0等式才成立.所以x=6,这样理解可以吗？解方程的时候，不能两边同时除以整式