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