题目

(x-5)/(x/7)+(x-2)/(x-4)=(x-3)/(x-5)+(x-4)/(x-6)

提问时间：2020-10-30

答案

(x-5)/(x-7)+(x-2)/(x-4)=(x-3)/(x-5)+(x-4)/(x-6)
(x-7+2)/(x-7)+(x-4+2)/(x-4)=(x-5+2)/(x-5)+(x-6+2)/(x-6)
1-2/(x-7)+1-2/(x-4)=1-2/(x-5)+1-2/(x-6)
2/(x-7)+2/(x-4)=2/(x-5)+2/(x-6)
1/(x-7)+1/(x-4)=1/(x-5)+1/(x-6)
(x-4+x-7)/(x-4)(x-7)=(x-6+x-5)/(x-5)(x-6)
(2x-11)/(x^2-11+28)=(2x-11)/(x^2-11+30)
因为x^2-11+28不等于x^2-11+30
分母不等,要等式成立只能分子为0
2x-11=0
x=11/2
经检验,x=11/2是原方程的解