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七年级数学填空题一般
题目
在有些情况下,不需要计算出结果也能把绝对值符号去掉,例如:6+7=6+7|6+7|=6+776=76|7-6|=7-667=76|6-7|=7-667=6+7|-6-7|=6+7.
(1)(1)根据上面的规律,把下列各式写成去掉绝对值符号的形式:
7+21=|7+21|=______;
12+0.8=|-\frac{1}{2}+0.8|=______;
2.823=______.|-2.8-\frac{2}{3}|=\_\_\_\_\_\_.
(2)(2)用合理的方法进行简便计算:9133+2120+4120(+2133)|-9\frac{1}{33}+2\frac{1}{20}|+|-4-\frac{1}{20}|-(+2\frac{1}{33})
(3)(3)用简单的方法计算:1312+1413+1514+...+1202012019+1202112020|\frac{1}{3}-\frac{1}{2}|+|\frac{1}{4}-\frac{1}{3}|+|\frac{1}{5}-\frac{1}{4}|+...+|\frac{1}{2020}-\frac{1}{2019}|+|\frac{1}{2021}-\frac{1}{2020}|.
知识点:绝对值(二)章节:第1章 有理数 / 1.2 有理数 / 1.2.4 绝对值

答案与解析

答案

(1)①7+21=7+21|7+21|=7+21.
故答案为:7+217+21
12+0.8=0.812|-\frac{1}{2}+0.8|=0.8-\frac{1}{2}
故答案为:0.8120.8-\frac{1}{2}
2.823=2.8+23|-2.8-\frac{2}{3}|=2.8+\frac{2}{3}
故答案为:2.8+232.8+\frac{2}{3}
(2)9133+2120+4120(+2133)(2)|-9\frac{1}{33}+2\frac{1}{20}|+|-4-\frac{1}{20}|-(+2\frac{1}{33})
=91332120+4+1202133=9\frac{1}{33}-2\frac{1}{20}+4+\frac{1}{20}-2\frac{1}{33}
=72+4=7-2+4
=9=9
(3)1312+1413+1514+...+1202012019+1202112020(3)|\frac{1}{3}-\frac{1}{2}|+|\frac{1}{4}-\frac{1}{3}|+|\frac{1}{5}-\frac{1}{4}|+...+|\frac{1}{2020}-\frac{1}{2019}|+|\frac{1}{2021}-\frac{1}{2020}|
=(1213)+(1314)+(1415)++(1201912020)+(1202012021)=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+…+(\frac{1}{2019}-\frac{1}{2020})+(\frac{1}{2020}-\frac{1}{2021})
=1212021=\frac{1}{2}-\frac{1}{2021}
=20194042=\frac{2019}{4042}.

解析

(1)①7+21=7+21|7+21|=7+21.
故答案为:7+217+21
12+0.8=0.812|-\frac{1}{2}+0.8|=0.8-\frac{1}{2}
故答案为:0.8120.8-\frac{1}{2}
2.823=2.8+23|-2.8-\frac{2}{3}|=2.8+\frac{2}{3}
故答案为:2.8+232.8+\frac{2}{3}
(2)9133+2120+4120(+2133)(2)|-9\frac{1}{33}+2\frac{1}{20}|+|-4-\frac{1}{20}|-(+2\frac{1}{33})
=91332120+4+1202133=9\frac{1}{33}-2\frac{1}{20}+4+\frac{1}{20}-2\frac{1}{33}
=72+4=7-2+4
=9=9
(3)1312+1413+1514+...+1202012019+1202112020(3)|\frac{1}{3}-\frac{1}{2}|+|\frac{1}{4}-\frac{1}{3}|+|\frac{1}{5}-\frac{1}{4}|+...+|\frac{1}{2020}-\frac{1}{2019}|+|\frac{1}{2021}-\frac{1}{2020}|
=(1213)+(1314)+(1415)++(1201912020)+(1202012021)=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+…+(\frac{1}{2019}-\frac{1}{2020})+(\frac{1}{2020}-\frac{1}{2021})
=1212021=\frac{1}{2}-\frac{1}{2021}
=20194042=\frac{2019}{4042}.

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