已知y＜根号x-1+根号1-x+二分之一,化简y-x分之根号x²-2xy+y²

题目

已知y＜根号x-1+根号1-x+二分之一,化简y-x分之根号x²-2xy+y²

答案

根号则x-1>=0,x>=1
1-x>=0,x0
所以原式=√(x-y)²/(y-x)
=|x-y|/(y-x)
=(x-y)/(y-x)
=-1