Question

若x,y都是实数,且满足y>√1/2-x +√x-1/2 +1 化简:|x-1|-√(x-1)² - √y

Answer 1

y>√1/2-x +√x-1/2 +1

先由根号得：X≥1/2,X≤1/2,所以X=1/2

√(y²-2y+1)/(y-1)=√(y-1)²/(y-1)=|y-1|

y>√1/2-x +√x-1/2 +1=1,y>1

|x-1|-√(x-1)² - √y²-2y+1/y-1=|x-1|-|x-1|-|y-1|

=-|y-1|=-(y-1)-y+1