解方程：[（x²-3x)/(x²-1）]+[(2x-1)/(x-1)]=0
问题描述：
解方程：[（x²-3x)/(x²-1）]+[(2x-1)/(x-1)]=0
答
[（x²-3x)/(x²-1）]+[(2x-1)/(x-1)]=0
(x²-3x)/(x-1)(x+1)+[(2x-1)(x+1)/(x-1)(x+1)]=0
[x²-3x+(2x-1)(x+1)]/(x-1)(x+1)]=0
[x²-3x+2x²+x-1]/(x-1)(x+1)]=0
(3x²-2x-1)/(x-1)(x+1)]=0
3x²-2x-1=0
(3x+1)(x-1)=0
x=-1/3或x=1
经x=-1/3是检验方程的解，x=1是增根
所以方程的解为：x=-1/3
答
x(x-3)+(x+1)(2x-1)=0
x²-3x+2x²+2x-x-1=0
3x²-4x-1=0
x=3分之2加减根号7
检验：x²-1不等于0
答
解析：由题意可知x≠1且x≠-1,那么：原方程两边同乘以x²-1可得：x²-3x+(2x-1)(x+1)=0x²-3x+2x²+x-1=0即3x²-2x-1=0(3x-1)(x+1)=0因为x≠-1,即x+1≠0,所以：解上述方程可得x=1/3即原方程的解...