Question

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

解方程1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)
要求用到规律：1/n(n+1)=1/n-1/(n+1)

Answer 1

1/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)
1/(x-3)-1/(x-2)-1/(x-4)+1/(x-1)+1/(x-2)-1/(x-1)=1/(x-4)
1/(x-3)-1/(x-4)=1/(x-4)
1/(x-3)=2/(x-4)
2(x-3)=x-4
2x-6=x-4
x=2
检验是增根
所以原分式方程无解

Answer 2

1/(x-2)(x-3)=1/(x-3)-1/(x-2), 这个把等号右边通分计算下之后就知道了
3/(x-1)(x-4)=1/(x-4)-1/(x-1),
1/(x-1)(x-2)=1/(x-2)-1/(x-1)
可以消掉中间项，方程就很简单了

Answer 3

/(x-2)(x-3)-3/(x-1)(x-4)+1/(x-1)(x-2)=1/(x-4)
1/(x-3)-1/(x-2)-1/(x-4)+1/(x-1)+1/(x-2)-1/(x-1)=1/(x-4)
1/(x-3)-1/(x-4)=1/(x-4)
1/(x-3)=2/(x-4)
2(x-3)=x-4
2x-6=x-4
x=2

Answer 4

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

Answer 5

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