试题

题目:
化简:
(1)(
3
-
2
)
0
-(-
1
2
)
-2
        
(2)(m-3n)-2·(2m-2n-3-2
(3)a-2b2·(-2a2b-2-2÷(a-4b2
(4)(2m2n-33(-mn-2-2
答案
解:(了)(
3
-
2
0-(-
2
-2
=了-4
=-3;

(2)(m-3n)-2·(2m-2n-3-2
=m7n-2·2-2m4n7
=
4
m7+4n-2+7
=
4
m了0n4

(3)a-2b2·(-2a2b-2-2÷(a-4b2
=a-2b2·(-2)-2a-4b4÷(a-4b2
=
4
a-2-4-(-4)b2+4-2
=
4
a-2b4
=
b4
4a2


(4)(2m2n-33(-mn-2-2
=23m7n-9(-m)-2n4
=8m7-2n-9+4
=8m4n-5
=
8m4
n5

解:(了)(
3
-
2
0-(-
2
-2
=了-4
=-3;

(2)(m-3n)-2·(2m-2n-3-2
=m7n-2·2-2m4n7
=
4
m7+4n-2+7
=
4
m了0n4

(3)a-2b2·(-2a2b-2-2÷(a-4b2
=a-2b2·(-2)-2a-4b4÷(a-4b2
=
4
a-2-4-(-4)b2+4-2
=
4
a-2b4
=
b4
4a2


(4)(2m2n-33(-mn-2-2
=23m7n-9(-m)-2n4
=8m7-2n-9+4
=8m4n-5
=
8m4
n5
考点梳理
负整数指数幂;零指数幂.
(1)根据任何非零数的零次幂等于1,负整数指数次幂等于正整数指数次幂的倒数进行计算即可得解;
(2)根据积的乘方的性质和同底数幂相乘,底数不变指数相加进行计算即可;
(3)根据积的乘方的性质,和同底数幂相乘,底数不变指数相加,然后利用负整数指数次幂等于正整数指数次幂的倒数计算;
(4)根据积的乘方的性质和同底数幂相乘,底数不变指数相减计算,再根据负整数指数次幂等于正整数指数次幂的倒数进行计算.
本题考查了负整数指数次幂等于正整数指数次幂的倒数,零指数幂,以及积的乘方的性质和同底数幂相乘,底数不变指数相加的性质,熟记性质是解题的关键,难点在于理清指数的变化.
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