Question

请简算这道题,并说明过程(1+0.12+0.23)×(0.12+0.23+0.34)--(1+0.12+0.23+0.34)×(0.12+0.23)

不要说什么设0.12+0.23=X,原式=0.34.我要小学生听的懂的,至少说一下关键的几步

Answer 1

(1+0.12+0.23)×(0.12+0.23+0.34)--(1+0.12+0.23+0.34)×(0.12+0.23)=(0.12+0.23+0.34)+(0.12+0.23)×(0.12+0.23+0.34)--(0.12+0.23)--(0.12+0.23+0.34)×(0.12+0.23)=(0.12+0.23+0.34)--(0.12+0.23)=0.34
第一步是把1和0.12+0.23分别与(0.12+0.23+0.34）相乘，减号后面是1和0.12+0.23+0.34分别与(0.12+0.23)相乘，而后可以约掉

Answer 2

(1+0.12+0.23)×(0.12+0.23+0.34)--(1+0.12+0.23+0.34)×(0.12+0.23)
＝[1+（0.12+0.23)]×[(0.12+0.23）+0.34)]-[(1+0.34+（0.12+0.23)]×(0.12+0.23)(把数归类0.12+0.23）
=(0.35*1)+0.35*0.35+0.34*0.35-[0.35*1+0.35*0.35+0.34*0.35]（利用乘法分配率）
=0

Answer 3

设0.12＋0.23＝a，0.12＋0.23+0.34＝b （1+0.12+0.23）*（0.12+0.23+0.34）-（1+0.12+0.23+0.34）*（0.12+0.23）
=(1+a)b-(1+b)a
=b-ab-a-ab
ab互相抵消：=b-a
=0.34 这答案很好吧！