Question

分式方程.x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)

Answer 1

(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)

[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/[(x-6)(x-2)]

[x^2-10x+9-x^2+6x+7]/(x^2-8x+7]=(x^2-10x+16-x^2+6x]/[x^2-8x+12]

(-4x+16)/(x^2-8x+7)=(-4x+16)/(x^2-8x+12)

所以-4x+16=0,

得：x=4

经检验为原方程的根