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数学
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(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)=______.
人气:274 ℃ 时间:2020-02-01 04:06:10
解答
设0.12+0.23=x,0.12+0.23+0.34=y.
则:(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)
=(1+x)y-(1+y)x,
=y+xy-x-xy,
=y-x,
=(0.12+0.23+0.34)-(0.12+0.23),
=0.12-0.12+0.23-0.23+0.34,
=0.34.
故答案为:0.34.
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简便运算;0.12+0.23+0.34+……+0.89=
(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)=_.
计算.要简便 (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+0.12+0.23+0.34)×(0.12+0.23)=_.
(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)=_.
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