解方程:1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)

来源:百度知道 编辑:UC知道 时间:2024/07/07 12:28:43
要写清楚

1/(x-7)-1/(x-5)=1/(x-6)-1/(x-4)
1/(x-7)+1/(x-4)=1/(x-6)+1/(x-5)
通分
(2x-11)/(x-7)(x-4)=(2x-11)/(x-6)(x-5)
(2x-11)[1/(x^2-11x+28)-1/(x^2-11x+30)]=0
因为x^2-11x+28不等于x^2-11x+30
所以1/(x^2-11x+28)-1/(x^2-11x+30)不等于0
所以所以2x-11=0
x=11/2
分式方程要检验
经检验,x=11/2是方程的解

设x-6=y
原式可换为:
1/(y-1) -1/(y+1) =1/y -1/(y+2)
2/(y²-1)=2/(y²+2y)
y²-1=y²+2y
y=-1/2
x=11/2