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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+767=76|6-7|=7-676=76|7-6|=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:
721=|7-21|=______;
120.8=|-\frac{1}{2}-0.8|=______;
717718=______:|\frac{7}{17}-\frac{7}{18}|=\_\_\_\_\_\_:
(2)(2)aa在数轴上的位置如图所示,则a2.5=______.|a-2.5|=\_\_\_\_\_\_.

A.a2.5A.a-2.5
B.2.5aB.2.5-a
C.a+2.5C.a+2.5
D.a2.5D.-a-2.5
(3)(3)利用上述介绍的方法计算或化简:
1512018+120181212+11009|\frac{1}{5}-\frac{1}{2018}|+|\frac{1}{2018}-\frac{1}{2}|-|-\frac{1}{2}|+\frac{1}{1009}
151a+1a1212+2(1a)|\frac{1}{5}-\frac{1}{a}|+|\frac{1}{a}-\frac{1}{2}|-|-\frac{1}{2}|+2(\frac{1}{a}),其中a>2a \gt 2.
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)①721=217|7-21|=21-7;②120.8=12+0.8|-\frac{1}{2}-0.8|=\frac{1}{2}+0.8;③717718=717718|\frac{7}{17}-\frac{7}{18}|=\frac{7}{17}-\frac{7}{18}
故答案为:①21721-7;②12+0.8\frac{1}{2}+0.8;③717718\frac{7}{17}-\frac{7}{18}
(2)(2)由数轴得:a<2.5a \lt 2.5
a2.5=2.5a|a-2.5|=2.5-a
故选:BB
(3)(3)利用上述介绍的方法计算或化简:
1512018+120181212+11009|\frac{1}{5}-\frac{1}{2018}|+|\frac{1}{2018}-\frac{1}{2}|-|-\frac{1}{2}|+\frac{1}{1009}
=1512018+121201812+11009=\frac{1}{5}-\frac{1}{2018}+\frac{1}{2}-\frac{1}{2018}-\frac{1}{2}+\frac{1}{1009}
=1511009+11009=\frac{1}{5}-\frac{1}{1009}+\frac{1}{1009}
=15=\frac{1}{5}
151a+1a1212+2(1a)|\frac{1}{5}-\frac{1}{a}|+|\frac{1}{a}-\frac{1}{2}|-|-\frac{1}{2}|+2(\frac{1}{a}),其中a>2a \gt 2.
2<a<52 \lt a \lt 5时,原式=1a15+121a12+2a=\frac{1}{a}-\frac{1}{5}+\frac{1}{2}-\frac{1}{a}-\frac{1}{2}+\frac{2}{a}
=15+2a=-\frac{1}{5}+\frac{2}{a}
=10a5a=\frac{10-a}{5a}
a5a\geqslant 5时,原式=151a+121a12+2a=\frac{1}{5}-\frac{1}{a}+\frac{1}{2}-\frac{1}{a}-\frac{1}{2}+\frac{2}{a}
=15=\frac{1}{5}.

解析

(1)①721=217|7-21|=21-7;②120.8=12+0.8|-\frac{1}{2}-0.8|=\frac{1}{2}+0.8;③717718=717718|\frac{7}{17}-\frac{7}{18}|=\frac{7}{17}-\frac{7}{18}
故答案为:①21721-7;②12+0.8\frac{1}{2}+0.8;③717718\frac{7}{17}-\frac{7}{18}
(2)(2)由数轴得:a<2.5a \lt 2.5
a2.5=2.5a|a-2.5|=2.5-a
故选:BB
(3)(3)利用上述介绍的方法计算或化简:
1512018+120181212+11009|\frac{1}{5}-\frac{1}{2018}|+|\frac{1}{2018}-\frac{1}{2}|-|-\frac{1}{2}|+\frac{1}{1009}
=1512018+121201812+11009=\frac{1}{5}-\frac{1}{2018}+\frac{1}{2}-\frac{1}{2018}-\frac{1}{2}+\frac{1}{1009}
=1511009+11009=\frac{1}{5}-\frac{1}{1009}+\frac{1}{1009}
=15=\frac{1}{5}
151a+1a1212+2(1a)|\frac{1}{5}-\frac{1}{a}|+|\frac{1}{a}-\frac{1}{2}|-|-\frac{1}{2}|+2(\frac{1}{a}),其中a>2a \gt 2.
2<a<52 \lt a \lt 5时,原式=1a15+121a12+2a=\frac{1}{a}-\frac{1}{5}+\frac{1}{2}-\frac{1}{a}-\frac{1}{2}+\frac{2}{a}
=15+2a=-\frac{1}{5}+\frac{2}{a}
=10a5a=\frac{10-a}{5a}
a5a\geqslant 5时,原式=151a+121a12+2a=\frac{1}{5}-\frac{1}{a}+\frac{1}{2}-\frac{1}{a}-\frac{1}{2}+\frac{2}{a}
=15=\frac{1}{5}.

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