题目

先化简在求值.（X＾2+2X分之X-2-X＾2+4X+4分之X-1）除X+2分之X-4.其中X＾2+2X...
先化简在求值.（X＾2+2X分之X-2-X＾2+4X+4分之X-1）除X+2分之X-4.其中X＾2+2X-1=0.

提问时间：2020-10-17

答案

原式={[(x-2)/(x²+2x)]-[(x-1)/(x²+4x+4)]}÷[(x-4)/(x+2)]
={(x-2)/[x(x+2)]-[(x-1)/(x+2)²]}×[(x+2)/(x-4)]
=(x-2)/[x(x+2)]×[(x+2)/(x-4)]-[(x-1)/(x+2)²]×[(x+2)/(x-4)]
=(x-2)/[x(x-4)]-(x-1)/[(x+2)(x-4)]
=[(x-2)(x+2)]/[x(x+2)(x-4)]-[x(x-1)]/[x(x+2)(x-4)]
=[(x-2)(x+2)-x(x-1)]/[x(x+2)(x-4)]
=(x²-4-x²+x)/[x(x+2)(x-4)]
=(x-4)/[x(x+2)(x-4)]
=1/[x(x+2)
=1/(x²+2x)
因为x²+2x-1=0
所以x²+2x=1
所以
原式=1/1=1