若a+m²=2004,b+m²=2003,c+m²=2002,且abc=24,求a∕bc+b∕ac+c∕ab‐1∕a‐1∕b‐1∕c的值
问题描述:
若a+m²=2004,b+m²=2003,c+m²=2002,且abc=24,求a∕bc+b∕ac+c∕ab‐1∕a‐1∕b‐1∕c的值
答
由a+m^2=2004,b+m^2=2003,c+m^2=2002可知A=C+2,B=C+1
把A=C+2,B=C+1代入abc=24,可得C^3+3*C^2+2C=24,易得C=2
所以A=4,B=3,C=2
所以a/bc+b/ca+c/ab-1/a-1/b-1/c=1/8