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设x=2006
1+2005^2+(2005^2/2006^2)
=1+(x-1)^2+(x-1)^2/x^2
=1+x^2-2x+1+1-2/x+1/x^2
=(x^2+2+1/x^2)-2*(x+1/x)+1
=(x+1/x)^2-2*(x+1/x)+1
=(x+1/x-1)^2
根号[1+2005^2+(2005^2/2006^2)]-1/2006
=x+1/x-1-1/x
=x-1
=2006-1
=2005
1+2005^2+(2005^2/2006^2)
=1+(x-1)^2+(x-1)^2/x^2
=1+x^2-2x+1+1-2/x+1/x^2
=(x^2+2+1/x^2)-2*(x+1/x)+1
=(x+1/x)^2-2*(x+1/x)+1
=(x+1/x-1)^2
根号[1+2005^2+(2005^2/2006^2)]-1/2006
=x+1/x-1-1/x
=x-1
=2006-1
=2005
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