a*a+b*b-2a+6b+10=0a^2-2a+1 + b^2+6b+9=0(a-1)^2 + (b+3)^2=0因为,(a-1)^2恒≥0 ,(b+3)^2恒≥0所以,(a-1)^2=0 ,(b+3)^2=0解得：a=1 ,b=-3a^2004 + 1/b =1^2004 + 1/(-3) = 1 - (1/3) =2/3