当x≠-1时,两边同时除以x+1,得:x-2=0,x=3,或者x=-1时,成立,所以解有两个!
X=3,X=-1
(x+1)(x-2)=x+1(x+1)(x-2)-(x+1)=0(x+1)(x-2-1)=0(x+1)(x-3)=0x=-1或x=3
(x+1)(x-2)-(x+1)=0(x+1)(x-2-1)=0x=3huo-1
(x+1)(x-2)=x+1(x+1)(x-2)-(x+1)=0(x+1)[(x-2)-1]=0(x+1)(x-3)=0x=-1,或x=3