-6000<-15<-12.6<-12<0.168<0.2<0.21<80<84<100.
∵5個(gè)有理數(shù)的兩兩乘積中有4個(gè)負(fù)數(shù)且沒(méi)有0,
∴這5個(gè)有理數(shù)中有1個(gè)負(fù)數(shù)和4個(gè)正數(shù),或者1個(gè)正數(shù)和4個(gè)負(fù)數(shù).
(1)若這5個(gè)有理數(shù)是1負(fù)4正,不妨設(shè)為x1<0<x2<x3<x4<x5,
則x1x5<x1x4<x1x3<x1x2<0<x2x3<x2x4<
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∴x1x5=-6000,x1x4=-15,x4x5=100,
三式相乘,得(x1x4x5)2=9×106,
又∵x1<0,x4>0,x5>0,
∴x1x4x5=-3000,
則x1=-30,x4=0.5,x5=200.
再由x1=-30,x1x2=-12,x1x3=-12.6,
得x2=0.4,x3=0.42.
經(jīng)檢驗(yàn)x1=-30,x2=0.4,x3=0.42,x4=0.5,x5=200滿(mǎn)足題意.
(2)若這5個(gè)有理數(shù)是4負(fù)1正.不妨設(shè)為:x1<x2<x3<x4<0<x5,
則x1x5<x2x5<x3x5<x4x5<0<x3x4<x2x4<
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∴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.
經(jīng)檢驗(yàn),x1=-200,x2=-0.5,x3=-0.42,x4=-0.4,x5=30滿(mǎn)足題意.
綜上可得:這5個(gè)有理數(shù)分別是-30,0.4,0.42,0.5,200或-200,-0.5,-0.42,-0.4,30.