题目内容

(1)(y22•y3÷(-y32
(2)(-
1
2
a3b)•(2bc2)3•(
1
2
a)2

(3)3x2•(-3xy)2-x2(x2y2-2x)
(4)(x+
1
2
)(x-
1
2
)-3x(
1
3
x-
2
3

(5)(4x+2)(x+3)-(2x-3)2
(6)[(x+y)2-(x-y)2]÷xy
分析:(1)先计算乘方,再计算乘除,即可化简;
(2)先去括号计算乘方,再计算乘除,即可化简;
(3)先计算乘方并去括号,再计算乘除,最后合并同类项,即可化简;
(4)先去括号,再合并同类项,即可化简;
(5)先去括号,再合并同类项,即可化简;
(6)先去小括号,再合并同类项,最后去大括号并计算乘除,即可化简.
解答:解:(1)原式=y4•y3÷y6
=y4+3-6
=y;
(2)原式=-
1
2
a3b
•23b3c6(
1
2
)
2
a2

=-a5b4c6

(3)原式=3x2•32x2y2-x4y2+2x3
=27x4y2-x4y2+2x3
=26x4y2+2x3

(4)原式=x2-
1
4
-x2+2x,
=2x-
1
4


(5)原式=4x2+12x+2x+6-4x2+12x-9,
=26x-3;

(6)原式=[x2+2xy+y2-x2+2xy-y2]÷xy,
=4xy÷xy,
=4.
点评:本题考查了同底数幂的计算以及合并同类项的计算.
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