(-1/42)÷(1/(6 )-3/14+2/3-2/7)

如题所述

第1个回答  推荐于2021-01-19
=(-1/42)÷(1/(6 )+2/3-(3/14+2/7))
=(-1/42)÷(1/(6 )+2/3-1/2)
=(-1/42)÷1/3
-1/14

(-1\/42)÷(1\/(6 )-3\/14+2\/3-2\/7)
=(-1\/42)÷(1\/(6 )+2\/3-(3\/14+2\/7))=(-1\/42)÷(1\/(6 )+2\/3-1\/2)=(-1\/42)÷1\/3 -1\/14

(-1\/42)除以(1\/6-3\/14+2\/3-2\/7)
(-1\/42)÷(1\/6+2\/3-3\/143-2\/7)=(-1\/42)÷(5\/6-3\/14-2\/7)=(-1\/42)÷[5\/6-(3\/14+2\/7)=(-1\/42)÷[5\/6-1\/2]=(-1\/42)÷1\/3 =(-1\/42)×3 =-1\/14

(负四十二分之一)÷(六分之一减十四分之三+三分之二减七分之二)?
先算括号里的,通分之后相加减,再根据除以一个数等于乘以它的倒数计算结果。

[-(1\/42)]\/(1\/6-3\/14+2\/3-2\/7)的简算
解法一:原式=(-1\/42)÷【1\/6+2\/3-(3\/14+2\/7)】=(-1\/42)÷1\/3=-1\/14 解法二:原式=(1\/6-3\/14+2\/3-2\/7)÷(-1\/42)=(1\/6-3\/14+2\/3-2\/7)×(-42)=-14 固原式=-1\/14 you knou?我很爱你的!

(-1\/42)除以(1\/6-3\/14+2\/3-2\/7)=多少?还有1又3分之一-7\/12+9\/20-11...
(-1\/42)除以(1\/6-3\/14+2\/3-2\/7)=-1÷(42×1\/6-42×3\/14+42×2\/3-42×2\/7)=-1÷(7-9+28-12)=-1÷14 =-1\/14 1又3分之一-7\/12+9\/20-11\/30+13\/42-15\/56+17\/72 =1+1\/3-1\/3-1\/4+1\/4+1\/5-1\/5-1\/6+1\/6+1\/7-1\/7-1\/8+1\/8+1\/9 =1+1\/...

用简便方法计算,(-1\/42)\/(1\/6-3\/14+2\/3-2\/7)
(-1\/42)\/(1\/6-3\/14+2\/3-2\/7)=(-1\/42)\/(1\/6+2\/3-3\/14-2\/7)=(-1\/42)\/[(1\/6+2\/3)-(3\/14+2\/7)]=(-1\/42)\/[5\/6-1\/2]=(-1\/42)\/[1\/3]=-1\/42*3 =-1\/14

选择合适的方法计算:(-1\/42)÷(1\/6-3\/14+2\/3-2\/7).
(-1\/42)÷(1\/6-3\/14+2\/3-2\/7)=(-1\/42)\/[1\/6+4\/6-(3\/14+2\/7)]=(-1\/42)\/[5\/6-1\/2)]=(-1\/42)\/(1\/3)=-1\/42*3 =-1\/14

用简便方法计算,(-1\/42)\/(1\/6-3\/14+2\/3-2\/7) 用原式的倒数这种方法
(-1\/42)\/(1\/6-3\/14+2\/3-2\/7)=(-1\/42)\/(1\/6+2\/3-3\/14-2\/7)=(-1\/42)\/[(1\/6+2\/3)-(3\/14+2\/7)]=(-1\/42)\/[5\/6-1\/2]=(-1\/42)\/[1\/3]=-1\/42*3 =-1\/14

(1)(-1\/42)÷(1\/6-3\/14+2\/3-2\/7)用简便方法 (2)(-2007\/5\/6)_百度...
(1)(-1\/42)÷(1\/6-3\/14+2\/3-2\/7)解:原式的倒数为(1\/6-3\/14+2\/3-2\/7)÷(-1\/42)=(1\/6-3\/14+2\/3-2\/7)×42 =1\/6×(-42)-3\/14×(-42)+2\/3×(-42)-2\/7×(-42)=-7-(-9)+(-28)-(-12)=2+(-28)-(-12)=-26-(-12)=-14 ...

负42分之一除六分之一减十四分之三加三分之二减七分之二等于?
解:-1\/42÷1\/6-3\/14+2\/3-2\/7 =-1\/42x6-3\/14-4\/14+2\/3 =-1\/7-7\/14+2\/3 =-2\/14-7\/14+2\/3 =-9\/14+2\/3 =-27\/42+28\/42 =1\/42(42分之1)亲,请【采纳答案】,您的采纳是我答题的动力,谢谢。

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