Question

a+x^2=2009,b+x^2=2010,c+x^2=2011,且abc=6030 求（a/bc）+（b/ac）+（c/ba）-1/a-1/b-1/c的值

Answer 1

a+x^2=2009,b+x^2=2010,c+x^2=2011,
得b-a=1,c-b=1.c-a=2
（a/bc）+（b/ac）+（c/ba）-1/a-1/b-1/c
=(a^2+b^2+c^2)/abc-(bc+ab+ac)/abc
=(a^2+b^2+c^2-bc-ab-ac)/abc
=2(a^2+b^2+c^2-bc-ab-ac)/2abc
=[(a-b)^2+(b-c)^2+(c-a)^2}/2abc
=(1+1+4)/2abc
=3/6030
=1/2010

Answer 2

化简原始得
原式=(a^2+b^2+c^2-bc-ac-ab)/abc
=[(a^2+b^2-2ab)/2+(b^2+c^2-2bc)/2+(c^2+a^2-2ca)/2]/abc
=[(a-b)^2+(b-c)^2+(c-a)^2]/2abc
=3/2/6030
=1/1005