解方程：（1）(2-x/x-3)+(1/3-x)=1 (2)(x-2/x+2)-(16/x^2-4)=1

来源：百度知道 编辑：UC知道 时间：2024/06/28 06:38:19

（1）
(2-x)/(x-3)+1/(3-x)=1
(2-x-1)/(x-3)=1
1-x=x-3
2x=4
x=2

(2)(x-2)/(x+2)-16/(x^2-4)=1
两边同乘以(x^2-4)
(x-2)^2-16=x^2-4
x^2-4x+4-16=x^2-4
4x=-8
x=-2
经检验得x=-2不符合，舍
所以无解

1
(2-x/x-3)+(1/3-x)=1
2-x-1=x-3
x=2

2
(x-2)^2-16=x^2-4
x^2-4x-12=x^2-4
x=-2,为增根，方程无解

（1）(2-x/x-3)+(1/3-x)=1
<=>(2-x)(x-3)-1/(x-3)=1.
<=>(2-x-1)/(x-3)-1=0
<=>[(1-x)-(x-3)]/(x-3)=0
<=>(4-2x)/(x-3)=0.
=>x=2.

(2)(x-2/x+2)-(16/x^2-4)=1
<=>(x-2)/(x+2)-16/(x^2-4)=1
<=>(x^2-4x+4)/(x^2-4)-16/(x^2-4)=1.
<=>(x^2-4x-12)/(x^2-4)-1=0.
<=>(x^2-4x-12-x^2+4)/(x^2-4)=0.
<=>(-4x-8)/(x^2-4)=0
<=>(x+2)/(x^2-4)=0
<=>(x+2)/[(x-2)(x+2)]=0
<=>不存在这样的x，那么，方程无解.