题目内容

计算:
(0.1+0.12+0.123+0.1234)×(0.12+0.123+0.1234+0.12345)-(0.1+0.12+0.123+0.1234+0.12345)×(0.12+0.123+0.1234)
思路点拨:这类题属于运用字母代换法计算的简便题,观察算式就发现括号部分算式是相同的,不妨将重复出现的算式用字母替换,这里不妨设:0.1+0.12+0.123+0.1234=A,0.12+0.123+0.1234=B,字母参与计算,计算就简便了.
考点:四则混合运算中的巧算
专题:计算问题(巧算速算)
分析:观察算式就发现括号部分算式是相同的,不妨将重复出现的算式用字母替换,这里不妨设:0.1+0.12+0.123+0.1234=A,0.12+0.123+0.1234=B,字母参与计算,计算就简便了.
解答: 解:设0.1+0.12+0.123+0.1234=A,0.12+0.123+0.1234=B
(0.1+0.12+0.123+0.1234)×(0.12+0.123+0.1234+0.12345)-(0.1+0.12+0.123+0.1234+0.12345)×(0.12+0.123+0.1234)
=A×(B+0.12345)-(A+0.12345)×B
=AB+0.12345A-AB-0.12345B
=012345×(A-B)
=0.12345×0.1
=0.012345.
点评:这类题属于运用字母代换法计算的简便,运用字母替换的方法,使计算简便.
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