Question

计算[根号1+2005^2+(2005^2/2006^2)]-(1/2006)

Answer 1

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