试题

题目:
计算下列各式:
(1)(-4
1
20
)×1.25×(-8);
(2)
5
6
×(-2.4)×
3
5

(3)(-14)×(-100)×(-6)×(0.01);
(4)9
18
19
×15;
(5)-100×
1
8
-0.125×35.5+14.5×(-12.5%);
(6)(1-2)×(2-3)×(3-4)×(4-5)×…(19-20).
答案
解:(1)(-4
1
20
)×1.25×(-8)
=(-
81
20
)×
5
4
×(-8)
=
81
20
×
5
4
×8
=
81
2


(2)
5
6
×(-2.4)×
3
5

=-
5
6
×2.4×
3
5

=-1.2;

(3)(-14)×(-100)×(-6)×(0.01)
=(-100)×(0.01)×(-14)×(-6)
=-1×84
=-84;

(4)9
18
19
×15
=(10-
1
19
)×15
=10×15-
1
19
×15
=150-
15
19

=149
4
19


(5)-100×
1
8
-0.125×35.5+14.5×(-12.5%)
=-100×0.125-0.125×35.5+14.5×(-0.125)
=0.125×(-100-35.5-14.5)
=0.125×(-150)
=-
75
4


(6)(1-2)×(2-3)×(3-4)×(4-5)×…×(19-20)
=(-1)×(-1)×(-1)×(-1)×…×(-1)
=-1.
解:(1)(-4
1
20
)×1.25×(-8)
=(-
81
20
)×
5
4
×(-8)
=
81
20
×
5
4
×8
=
81
2


(2)
5
6
×(-2.4)×
3
5

=-
5
6
×2.4×
3
5

=-1.2;

(3)(-14)×(-100)×(-6)×(0.01)
=(-100)×(0.01)×(-14)×(-6)
=-1×84
=-84;

(4)9
18
19
×15
=(10-
1
19
)×15
=10×15-
1
19
×15
=150-
15
19

=149
4
19


(5)-100×
1
8
-0.125×35.5+14.5×(-12.5%)
=-100×0.125-0.125×35.5+14.5×(-0.125)
=0.125×(-100-35.5-14.5)
=0.125×(-150)
=-
75
4


(6)(1-2)×(2-3)×(3-4)×(4-5)×…×(19-20)
=(-1)×(-1)×(-1)×(-1)×…×(-1)
=-1.
考点梳理
有理数的乘法.
(1)把带分数化为假分数,小数化为分数,然后根据有理数的乘法法则进行计算即可得解;
(2)根据有理数的乘法运算法则进行计算即可得解;
(3)利用乘法交换结合律进行计算即可得解;
(4)把9
18
19
写成(10-
1
19
),然后利用乘法分配律进行计算即可得解;
(5)逆运用乘法分配律进行计算即可得解;
(6)先算小括号里面的,再根据有理数的乘法运算法则进行计算即可得解.
本题考查了有理数的乘法,利用运算定律计算可以使计算更加简便,计算时要注意运算符号的处理.
计算题.
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