答
将5个有理数两两的乘积由小到大排列:
-6000<-15<-12.6<-12<0.168<0.2<0.21<80<84<100.
∵5个有理数的两两乘积中有4个负数且没有0,
∴这5个有理数中有1个负数和4个正数,或者1个正数和4个负数.
(1)若这5个有理数是1负4正,不妨设为x1<0<x2<x3<x4<x5,
则x1x5<x1x4<x1x3<x1x2<0<x2x3<x2x4<<x3x5<x4x5,(其中x2x5和x3x4的大小关系暂时还不能断定),
∴x1x5=-6000,x1x4=-15,x4x5=100,
三式相乘,得(x1x4x5)2=9×106,
又∵x1<0,x4>0,x5>0,
∴x1x4x5=-3000,
则x1=-30,x4=0.5,x5=200.
再由x1=-30,x1x2=-12,x1x3=-12.6,
得x2=0.4,x3=0.42.
经检验x1=-30,x2=0.4,x3=0.42,x4=0.5,x5=200满足题意.
(2)若这5个有理数是4负1正.不妨设为:x1<x2<x3<x4<0<x5,
则x1x5<x2x5<x3x5<x4x5<0<x3x4<x2x4<<x1x3<x1x2,(其中x1x4和x2x3的大小关系暂时还不能断定),
∴x1x5=-6000,x2x5=-15,x1x2=100,
三式相乘,得(x1x2x5)2=9×106,
又∵x1<0,x2<0,x5>0,
解得x1x2x5=3000,
∴x1=-200,x2=-0.5,x5=30,
再由x5=30,x3x5=-12.6,x4x5=-12,
得x3=-0.42,x4=-0.4.
经检验,x1=-200,x2=-0.5,x3=-0.42,x4=-0.4,x5=30满足题意.
综上可得:这5个有理数分别是-30,0.4,0.42,0.5,200或-200,-0.5,-0.42,-0.4,30.
答案解析:首先将将5个有理数两两的乘积由小到大排列,由5个有理数的两两乘积中有4个负数且没有0,可得这5个有理数中有1个负数和4个正数,或者1个正数和4个负数.再分别从若这5个有理数是1负4正,不妨设为x1<0<x2<x3<x4<x5,可得x1x5<x1x4<x1x3<x1x2<0<x2x3<x2x4<<x3x5<x4x5,(其中x2x5和x3x4的大小关系暂时还不能断定),若这5个有理数是4负1正.不妨设为:x1<x2<x3<x4<0<x5,则x1x5<x2x5<x3x5<x4x5<0<x3x4<x2x4<<x1x3<x1x2,(其中x1x4和x2x3的大小关系暂时还不能断定),去分析求解即可求得答案.
考试点:有理数无理数的概念与运算.
知识点:此题考查了有理数的概念与运算.此题难度较大,注意根据题意得到这5个有理数中有1个负数和4个正数,或者1个正数和4个负数,然后分类讨论求解是解此题的关键.