Question

已知a+x²=2009,b+x²=2010,c+x²=2011,且abc=6030,求a/

Answer 1

a+1=b；b+1=c；a+2=c

a/bc+b/ca+c/ab-1/a-1/b-1/c

=(a^2+b^2+c^2-bc-ac-ab)/abc

=(a^2+b^2+c^2-bc-ac-ab)/6030

=(a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac)/(6030*2)

=((a-b)^2+(a-c)^2+(b-c)^2)/(6030*2)

=(1+4+1)/(6030*2)=1/2010