计算题:(x的平方-x-2分之x-1）的平方除以2-x分之x的平方-2x+1除以（x的平方+x分之1）的平方

原式=[(x-1)/(x²-x-2)]²÷[(x²-2x+1)/(2-x)]÷[1/(x²+x)]²
=[(x-1)²/(x²-x-2)²]×[(2-x)/(x²-2x+1)]×(x²+x)²
={(x-1)²/[(x+1)(x-2)]²}×[-(x-2)/(x-1)²]×[x(x+1)]²
={(x-1)²/[(x+1)²(x-2)²]}×[-(x-2)/(x-1)²]×[x²(x+1)²]
={-1/[(x+1)²(x-2)]}×[x²(x+1)²]
=-x²/(x-2)
结果等于:负(x-2)分之x²