已知：a+x2=2015，b+x2=2016，c+x2=2017，且abc=12，则abc+cab+bac-1a-1b-1c=．

Question

已知：a+x²=2015，b+x²=2016，c+x²=2017，且abc=12，则 (a)/(bc)+ (c)/(ab)+ (b)/(ac)- (1)/(a)- (1)/(b)- (1)/(c)=___．

Answer 1

由题意得：

a+x2=2015①

b+x2=2016②

c+x2=2017③

①-②得：a-b=-1
①-③得：a-c=-2
②-③得：b-c=-1
∴(a)/(bc)+(c)/(ab)+(b)/(ac)-(1)/(a)-(1)/(b)-(1)/(c)
=(a2+c2+b2-bc-ac-ab)/(abc)
=((a-c)2+(a-b)2+(b-c)2)/(2abc)
=((-2)2+(-1)2+(-1)2)/(2×12)
=(1)/(4)=0.25
故答案为：0.25