解两个分式方程
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/17 19:08:41
解两个分式方程
![](http://img.wesiedu.com/upload/6/5d/65dc922527ca6d53a2f3fe495e9e594a.jpg)
![](http://img.wesiedu.com/upload/6/5d/65dc922527ca6d53a2f3fe495e9e594a.jpg)
![解两个分式方程](/uploads/image/z/19261238-14-8.jpg?t=%E8%A7%A3%E4%B8%A4%E4%B8%AA%E5%88%86%E5%BC%8F%E6%96%B9%E7%A8%8B)
1) 去分母: (3x-1)(x-1)+(2-x)(x+1)=(x^2-1)+2
3x^2-4x+1+2+x-x^2=x^2+1
x^2-3x+2=0
(x-1)(x-2)=0
x=1(增根,舍去)或x=2
故原方程的根为x=2
2)合并:
[1/(x+1)+1/(x-1)]+[1/(x+2)+1/(x-2)]=0
2x/(x^2-1)+2x/(x^2-4)=0
故x=0或1/(x^2-1)+1/(x^2-4)=0
后者去分母得:x^2-4+x^2-1=0,即x^2=5/2,得:x=√10/2或-√10/2
经检验原方程有这3个根:x=0, √10/2,-√10/2
3x^2-4x+1+2+x-x^2=x^2+1
x^2-3x+2=0
(x-1)(x-2)=0
x=1(增根,舍去)或x=2
故原方程的根为x=2
2)合并:
[1/(x+1)+1/(x-1)]+[1/(x+2)+1/(x-2)]=0
2x/(x^2-1)+2x/(x^2-4)=0
故x=0或1/(x^2-1)+1/(x^2-4)=0
后者去分母得:x^2-4+x^2-1=0,即x^2=5/2,得:x=√10/2或-√10/2
经检验原方程有这3个根:x=0, √10/2,-√10/2