题目是不是这样:[(x²-x-2)/(x²-4x+4)-x/(x²-2x)]·(x-4/x)
如果是的话,解题如下
原式=[(x-2)(x+1)/(x-2)²-x/(x²-2x)]·(x-4/x)
=[(x+1)/(x-2)-1/(x-2)]·(x-4/x)
=[x/(x-2)]·(x-4/x)
=(x-4)/(x-2)
将x=-1代入,原式=5/3
:(x²-x-2)/x²-4x+4-x/x²-2x)·(x-4/x)
=((x-2)(x+1)/(x-2)²-x/x(x-2))(x-4/x)
=((x+1)/(x-2)-1/(x-2))((x²-4)/x)
=(x/(x-2))(x+2)(x-2)/x
=x+2
=2-1
=1
你的括号有没有问题