试题

题目:
(1)(1×1a1a)÷
(-q×1a7
(-q×1a7
=-r×1a7
(r)
r7
r
a-1x-r
r7
r
a-1x-r
·(-
r
7
arxr)=-7a;
(q)
(ab-bn+rcn)
(ab-bn+rcn)
÷n=a-b+rc;
(4)(qxqyr+x4yr-
1
r
xy
1
r
xy
)÷
1
r
xy=
1xry
1xry
+
rxqy
rxqy
-1.
答案
(-q×1a7

r7
r
a-1x-r

(ab-bn+rcn)

1
r
xy

1xry

rxqy

解:(1)(6×1010)÷(-4×105)=-1×105
(1)(
15
1
a-1x-1)·(-
1
5
a1x1)=-5a;
(4)(ab-bn+1cn)÷n=a-b+1c;
(4)(4x4y1+x4y1-
1
1
xy)÷
1
1
xy=6x1y+1x4y-1.
故答案为:(1)-4×105;(1)
15
1
a-1x-1;(4)ab-bn+1cn;(4)
1
1
xy;6x1y;1x4y.
考点梳理
整式的混合运算.
各式利用运算法则计算即可得到结果.
此题考查了整式的混合运算,熟练掌握运算法则是解本题的关键.
计算题.
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