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Defining $$x*y=\dfrac{\sqrt{x^2+3xy+y^2-2x-2y+4}}{xy+4},$$ compute $$((\cdots((2007*2006)*2005)*\cdots)*1).$$

I think this must find good $x,y$ such that $xy=const$

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1 Answer 1

up vote 10 down vote accepted

Hint: $\sqrt{x^2 + 3xy + y^2 - 2x - 2y + 4} = \sqrt{(x+y)^2 + (x-2)(y-2)}$

What happens when $y = 2$?

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it's very nice,Thank you –  math110 May 5 '13 at 13:37
    
(+1) nice observation. –  achille hui May 5 '13 at 13:39

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