试题

题目:
(1)(2
1
4
2×42
(2)[(
1
2
2]3×(233
(3)(-0.125)12×(-1
2
3
7×(-8)13×(-
3
5
9
(4)-82003×(0.125)2002+(0.25)17×417
答案
(1)(2
1
4
2×42
=(
9
4
2×42
=81;

(2)[(
1
2
2]3×(233
=(
1
2
6×29
=(
1
2
×2)6×23
=23
=8;

(3)(-0.125)12×(-1
2
3
7×(-8)13×(-
3
5
9
=(-
1
8
12×(-
5
3
7×(-8)13×(-
3
5
9
=-(
1
8
12×813×(
5
3
7×(
3
5
9
=-(
1
8
×8)12×8×(
5
3
×
3
5
7×(
3
5
2
=-8×
9
25

=-
72
25


(4)-82003×(0.125)2002+(0.25)17×417
=-8×82002×(
1
8
2002+(
1
4
17×417
=-(8×
1
8
2002×8+(
1
4
×4)17
=-8+1,
=-7;
(1)(2
1
4
2×42
=(
9
4
2×42
=81;

(2)[(
1
2
2]3×(233
=(
1
2
6×29
=(
1
2
×2)6×23
=23
=8;

(3)(-0.125)12×(-1
2
3
7×(-8)13×(-
3
5
9
=(-
1
8
12×(-
5
3
7×(-8)13×(-
3
5
9
=-(
1
8
12×813×(
5
3
7×(
3
5
9
=-(
1
8
×8)12×8×(
5
3
×
3
5
7×(
3
5
2
=-8×
9
25

=-
72
25


(4)-82003×(0.125)2002+(0.25)17×417
=-8×82002×(
1
8
2002+(
1
4
17×417
=-(8×
1
8
2002×8+(
1
4
×4)17
=-8+1,
=-7;
考点梳理
整式的混合运算;幂的乘方与积的乘方.
(1)根据积的乘方的性质的逆运用求解;
(2)先整理出同指数幂的乘法,再利用积的乘方的性质的逆运用求解;
(3)根据乘法交换律和结合律,把第一三项,二四项先整理成同指数幂的运算,然后根据积的乘方性质的逆运用求解;
(4)分别利用积的乘方的性质的逆运用进行计算即可.
本题主要考查积的乘方的性质的逆运用,熟练掌握运算性质,整理成同底数幂的乘法是解题的关键.
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