求方程解x-5/x-4+x-9/x-8=x-8/x-7+x-6/x-5

[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]
1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)
(x-4+x-8)/(x-4)(x-8)=(x-7+x-5)/(x-7)(x-5)
(2x-12)/(x^2-12+32)=(2x-12)/(x^2-12+35)
(2x-12)[1/(x^2-12+32)-1/(x^2-12+35)]=0
x^2-12+32≠x^2-12+35
所以1/(x^2-12+32)-1/(x^2-12+35)≠0
所以2x-12=0
x=6
分式方程要检验
经检验
x=6是方程的解