Question

一道数学题 已知正数a、b、c、d、e、f满足:

bcdef/a =4
acdef/b =9
abdef/c =16
abcef/d =1/4
abcdf/e =1/9
abcde/f =1/16
则(a+c+e)-(b+d+f)=?

Answer 1

六个式子相加乘可得　１／abcdef＝１　即abcdef＝１
　将bcdef看做一个数　即可推出a＝１／２
同理可根据其他式子推出　b＝１／３　　
c＝１／４　　d＝２　　e＝３　f＝４
那么（a＋c＋e）－（b＋d＋f）＝－３１／１２

Answer 2

因为bcdef/a =4，acdef/b =9，abdef/c =16，abcef/d =1/4，abcdf/e =1/9，abcde/f =1/16
所以（bcdef/a）×（acdef/b）×（abdef/c )×（abcef/d）×（abcdf/e）×（abcde/f）
=[(a^5)×(b^5)×(c^5)×(d^5)×(e^5)×(f^5)] / (abcdef)
=(a^4)×(b^4)×(c^4)×(d^4)×(e^4)×(f^4)=1
所以 (abcdef)^4=1
abcdef=±1
又因为a、b、c、d、e、f为正数
所以abcdef=1
又因为 bcdef/a =4,所以 (bcdef/a)×1=abcdef/a�0�5 =4
又因为acdef/b =9,所以(acdef/b)×1=abcdef/b�0�5 =9
又因为abdef/c =16,所以(abdef/c)×1=abcdef/c�0�5 =16
又因为abcef/d =1/4,所以(abcef/d)×1=abcdef/d�0�5 =1/4
又因为abcdf/e =1/9,所以(abcdf/e)×1=abcdef/e�0�5 =1/9
又因为abcde/f =1/16,所以(abcde/f)×1=abcdef/f�0�5 =1/16
所以1/a�0�5=4, 1/b�0�5=9, 1/c�0�5=16, 1/d�0�5=1/4, 1/e�0�5=1/9, 1/f�0�5=1/16
所以a=1/2 ,b=1/3, c=1/4, d=2,e=3, f=4,
(a+c+e)-(b+d+f)=[(1/2)+(1/4)+3]-[(1/3)+2+4]=-31/12

Answer 3

六个式子两边分别相乘得　(abcdef)^4＝１　abcdef＝１
　bcdef/a=abcdef/a^2=1/a^2=4　a＝1/2
同样可得　b＝1/3　
c＝1/4　　d＝２　　e＝３　f＝４
(a+c+e)-(b+d+f)=-31/12

Answer 4

33