题目内容
填空:
(1)(-8)12÷(-8)25= ;
(2)x3÷x2= ;
(3)(23×26)2÷28= ;
(4)(x5÷x2)÷(x7÷x6)= ;
(5)已知3n=2,3m=5,则32m-3n= .
(1)(-8)12÷(-8)25=
(2)x3÷x2=
(3)(23×26)2÷28=
(4)(x5÷x2)÷(x7÷x6)=
(5)已知3n=2,3m=5,则32m-3n=
考点:同底数幂的除法,幂的乘方与积的乘方
专题:
分析:分别根据同底数幂相除,底数不变指数相减;同底数幂相乘,底数不变指数相加进行计算即可得解.
解答:解:(1)(-8)12÷(-8)25
=(-8)12-25
=(-8)-13
=-
;
(2)x3÷x2=x3-2=x;
(3)(23×26)2÷28
=(23+6)2÷28
=218÷28
=210;
(4)(x5÷x2)÷(x7÷x6)
=x5-2÷x7-6
=x3÷x
=x2;
(5)∵3n=2,3m=5
∴32m-3n=(3n)2÷(3m)3
=22÷53
=
.
故答案为:-
,x,210,x2,
.
=(-8)12-25
=(-8)-13
=-
| 1 |
| 813 |
(2)x3÷x2=x3-2=x;
(3)(23×26)2÷28
=(23+6)2÷28
=218÷28
=210;
(4)(x5÷x2)÷(x7÷x6)
=x5-2÷x7-6
=x3÷x
=x2;
(5)∵3n=2,3m=5
∴32m-3n=(3n)2÷(3m)3
=22÷53
=
| 4 |
| 125 |
故答案为:-
| 1 |
| 813 |
| 4 |
| 125 |
点评:本题考查了同底数幂的乘法,同底数幂的除法的性质,熟记性质并理清指数的变化是解题的关键.
练习册系列答案
相关题目