试题

题目:
计算:(1)(如02·(04&nb0p;2-(-052·(022
(2)(-202b4+08·(-2b4
(如)(-1)2229+(-
1
2
-2-(π-1)2-|-如|
(4)-2122×2.5122×(-1)999
(5)(p-q)&nb0p;6÷[(p-q)&nb0p;2·(q-p)&nb0p;如]
(6)已知如×9m×27m=如16,求m的值
答案
解:(1)(3a33·(a4 3-(-a53·(a33
=37a6·a8-a10·a4
=37a14-a14
=36a14

(3)(-3a3b34+a8·(-3b43
=16a8b13-8a8b13
=8a8b13

(3)(-1)3009+(-
1
3
-3-(π-1)0-|-3|,
=-1+4-1-3,
=-1;

(4)-3100×0.5100×(-1)999
=-(3×0.5)100×(-1),
=1;

(5)(p-q) 6÷[(p-q) 3·(q-p) 3],
=(p-q) 6÷[-(p-q) 5],
=-(p-q),
=q-p;

(6)∵3×9m×37m=316
∴31+3m+3m=316
∴1+3m+3m=16,
解得m=3.
解:(1)(3a33·(a4 3-(-a53·(a33
=37a6·a8-a10·a4
=37a14-a14
=36a14

(3)(-3a3b34+a8·(-3b43
=16a8b13-8a8b13
=8a8b13

(3)(-1)3009+(-
1
3
-3-(π-1)0-|-3|,
=-1+4-1-3,
=-1;

(4)-3100×0.5100×(-1)999
=-(3×0.5)100×(-1),
=1;

(5)(p-q) 6÷[(p-q) 3·(q-p) 3],
=(p-q) 6÷[-(p-q) 5],
=-(p-q),
=q-p;

(6)∵3×9m×37m=316
∴31+3m+3m=316
∴1+3m+3m=16,
解得m=3.
考点梳理
整式的混合运算.
(1)(2)先乘方,再乘除,最后合并同类项;(3)先乘方,再合并;(4)利用积的乘方法则计算;(5)把p-q看作整体,做同底数幂的乘除法运算;(6)将左边化为3为底数的同底数幂进行运算.
本题考查了整式的混合运算.关键是按照混合运算的顺序进行,同时注意运算法则的灵活运用.
计算题.
找相似题