a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
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![a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?](/uploads/image/z/745571-11-1.jpg?t=a%2Bm%28%E5%B9%B3%E6%96%B9%29%3D2004%2Cb%2Bm%28%E5%B9%B3%E6%96%B9%29%3D2003%2Cc%2Bm%28%E5%B9%B3%E6%96%B9%29%3D2002%2C%E4%B8%94abc%3D24%2C%E5%88%99a%2Fbc%2Bb%2Fca%2Bc%2Fab-1%2Fa-1%2Fb-1%2Fc%3D%3F)
a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
a+m(平方)=2004,b+m(平方)=2003,c+m(平方)=2002
可得 a-b=1 b-c=1 a-c=2
a/bc+b/ca+c/ab-1/a-1/b-1/c
=(a^2+b^2+c^2-bc-ac-ab)/abc
=1/2 (2a^2 + 2b^2 + 2c^2- 2bc - 2ac - 2ab)/abc
=1/2((a^2-2ac+c^2)+(b^2--2bc +c^2)+(a^2-2ab+b^2))/abc
=1/2((a-c)^2+(b-c)^2+(a-b)^2)/abc
=1/8
有前三个式子的,a=b+1,c=b-1,通分最后一个式子=(a2+b2+c2-bc-ac-ab)/abc=[a(a-b)+b(b-c)+c(c-a)]/24=(a+b-2c)/24=[b+1+b-2(b-1)]/24=3/24=1/8,就是这样了,搞了半天,麻烦给个好评哈,亲
a+m²=2004,b+m²=2003,c+m²=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
m²=2004-a=2003-b=2002-c,故2-a=1-b=-c,
即有a-b=1......(1);a-c=2........(2);b-c=1.........(3);abc=24...........
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a+m²=2004,b+m²=2003,c+m²=2002,且abc=24,则a/bc+b/ca+c/ab-1/a-1/b-1/c=?
m²=2004-a=2003-b=2002-c,故2-a=1-b=-c,
即有a-b=1......(1);a-c=2........(2);b-c=1.........(3);abc=24........(4)
由(1)(2)(3)(4)得a=4,b=3,c=2;
∴a/bc+b/ca+c/ab-1/a-1/b-1/c=4/6+3/8+2/12-1/4-1/3-1/2=1/8
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