Question

若x,y为实数,且y＜根号x-1+根号1-x+2分之1,化简：1分之y-1根号1-2y+y².

Answer 1

y＜√(x-1) + √(1-x) + 1/2
x-1≥0,
1-x≥0,即x-1≤0
∴x-1=0,x=1
∴y＜√(x-1) + √(1-x) + 1/2 = 1/2
∴1-y＞0
∴|1-y|/(y-1) = (1-y)/(y-1) = -1