当x>2时，使不等式x+1/(x-2)>=a恒成立的实数a的取值范围是？

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x>2
x-2>0
所以x+1/(x-2)
=(x-2)+1/(x-2)+2
>悉衡=2根号[(x-2)*1/(x-2)]+2
=2+2
=4
所世陆洞搜枯以a<=4

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x+1/(x-2)>=a
(x-2)+1/(x-2)+2>=a
(x-2)+1/(x-2)+2>差盯=2√[(x-2)*1/卜庆巧(x-2)]+2=4
a<型键=4

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