已知x,y满足y

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已知x,y满足y已知x,y满足y已知x,y满足yy＜√(x-1)+√(1-x)+1/2x-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＞

已知x,y满足y
已知x,y满足y

已知x,y满足y
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

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