先化简,再求值:[(x-2/x²+2x)-(x-1/x²+4x+4)]÷﹙x-4/x+2﹚,其中x²+2x-1=0.
问题描述:
先化简,再求值:[(x-2/x²+2x)-(x-1/x²+4x+4)]÷﹙x-4/x+2﹚,其中x²+2x-1=0.
答
[(x-2/x²+2x)-(x-1/x²+4x+4)]÷﹙x-4/x+2﹚
=[(x-2/x(x+2)-(x-1/(x+2)²]÷﹙x-4/x+2﹚
=[(x-2)(x+2)/x(x+2)²-x(x-1)/x(x+2)²]÷﹙x-4/x+2﹚
={[(x-2)(x+2)-x(x-1)]/x(x+2)²}÷﹙x-4/x+2﹚
={(x-4)/x(x+2)²}*[﹙x+2)/(x-4)]
=1/x(x+2)
=1/(x²+2x)
因x²+2x-1=0,所以x²+2x=1
所以原式=1/1=1