Question

解：(x-2分之x)+(x-7分之x-9)=(x-1分之x+1)+(x-6分之x-8)

Answer 1

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)
(x-2+2)/(x-2)+(x-7-2)/(x-7)=(x-1+2)/(x-1)+(x-6-2)/(x-6)
1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)
1/(x-2)+1-1/(x-7)=1/(x-1)-1/(x-6)
[x-7-(x-2)]/[(x-1)(x-2)]=[x-6-(x-1)][(x-6)(x-7)]
-5/[(x-1)(x-2)]=-5[(x-6)(x-7)]
(x-1)(x-2)=(x-6)(x-7)
x^2-3x+2=x^2-13x+42
10x=40
x=4
经验证，x=4不是增根

追问

最简公分母是？

追答

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)
x/(x-2)-(x-8)/(x-6)=(x+1)/(x-1)-(x-9)/(x-7)
(x^2-6x-x^2+10x-16)/(x^2-8x+12)=(x^2-6x-7-x^2+10x-9)/(x^2-8x+7)
两式分子相等,分母不等
不需要求最简公分母
由于分子相等,分母不等‘

所以x^2-6x-x^2+10x-16=0
解得x=4