,y∈R+ (23 19:53:18)

题目

,y∈R+ (23 19:53:18)
1、设x,y∈R+且xy-(x+y)=1,则
A.x+y≥2[(根号2)+1]
B.xy≤(根号2)+1
C.x+y≤[(根号2)+1] 2
D.xy≥2[(根号2)+1]
 
 

答案

xy-(x+y)=1 xy-x=y+1 x=(y+1)/(y-1) x=1+2/(y-1) 因为x>0,y>0 所以y+1>0,那么y-1>0.x+y=1+2/(y-1)+y=2+2/(y-1)+(y-1)>=2+2√2 所以x+y最小值是2+2√2 PS：运用的是均值不等式 当y-1>0时,当2/(y-1)与（y-1)相等时,可...