,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)+1C.x+y≤[(根号2)+1] 2D.xy≥2[(根号2)+1]  

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,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)+1C.x+y≤[(根号2)+1] 2D.xy≥2[(根号2)+1]  
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,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)+1C.x+y≤[(根号2)+1] 2D.xy≥2[(根号2)+1]  
,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]
 
 

,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)+1C.x+y≤[(根号2)+1] 2D.xy≥2[(根号2)+1]  
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,
可以取最小值2+2√2
A.x+y≥2[(根号2)+1]

xy大于等于[（根号2）+1]；

xy-x-y=1 得出x(y-1)=y+1 x=(y+1)/(y-1) 因为x>0 且y>0 所以y+1>0 故y-1只能大于0，有x=(y+1)/(y-1)=1+2/(y-1)
所以x+y=y+1+2/(y-1)=(y-1)+2/(y-1)+2 易得x+y>=2+2根号2=2【根号（2）+1】
选A

xy-(x+y)=1===>(x+y)+1=xy《[(x+y)/2]^2===>(x+y)^2》4(x+y)+4===>[(x+y)-2]^2》8===>|(x+y)-2|》2√2===>(x+y)》2[1+√2]或0<（x+y)《2(1-√2）<0(舍去）===>x+y》2[1+√2].选A.

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