求解分式方程.(6x+12)/(x^2+4x+4)-(x^2-4)/(x^2-4x+4)+x^2/(x^2-4)=0_数学解答

求解分式方程.(6x+12)/(x^2+4x+4)-(x^2-4)/(x^2-4x+4)+x^2/(x^2-4)=0

人气：491 ℃　时间：2020-05-31 16:31:06

解答

方程化为：6/(x+2)-(x+2)/(x-2)+x^2/(x+2)(x-2)=0
两边同时乘以(x-2)(x+2)
6(x-2)-(x+2)^2+x^2=0
6x-12-4x-4=0
2x=16
x=8
经检验为原方程的根.