已知，1a+1b＝16，1b+1c＝19，1a+1c＝115，则abcab+bc+ac=．

Question

已知，(1)/(a)+(1)/(b)＝(1)/(6)，(1)/(b)+(1)/(c)＝(1)/(9)，(1)/(a)+(1)/(c)＝(1)/(15)，则(abc)/(ab+bc+ac)=______．

Answer 1

由已知得，
(1)/(a)+(1)/(b)＝(1)/(6)⇒(ac+bc)/(abc)=(1)/(6)，
(1)/(b)+(1)/(c)＝(1)/(9)⇒(ab+ac)/(abc)=(1)/(9)，
(1)/(a)+(1)/(c)＝(1)/(15)⇒(ab+bc)/(abc)=(1)/(15)，
∴(2(ab+bc+ac))/(abc)=(31)/(90)，
∴(abc)/(ab+bc+ac)=(45)/(31)．
故答案为：(45)/(31)．