试题

题目:
用心算一算:
(图)2a5·(-a)2-(-a23·(-7a);
(2)(4x2y+5xy-7x)-(5x2y+4xy+x);
(3)(
2
x2y-2xy+y2)·3xy;
(4)(4x3y-6x2y2+图2xy3)÷(2xy);
(5)化简求值(x+2y)2-(x+y)(x-y),其ox=2,y=
2

答案
解:(1)2az·(-a)2-(-a23·(-7a),
=2a7-7a7
=-za7

(2)(4b2y+zby-7b)-(zb2y+4by+b),
=-b2y+by-8b;

(3)(
1
2
b2y-2by+y2)·3by
=
3
2
b3y2-2b2y2+3by3

(4)(4b3y-2b2y2+12by3)÷(2by),
=2b2-3by+2y2

(z)(b+2y)2-(b+y)(b-y),
=b2+4by+4y2-b2+y2
=4by+zy2
当b=2,y=
1
2
时,原式=4×2×
1
2
+z×
1
4
=
21
4

解:(1)2az·(-a)2-(-a23·(-7a),
=2a7-7a7
=-za7

(2)(4b2y+zby-7b)-(zb2y+4by+b),
=-b2y+by-8b;

(3)(
1
2
b2y-2by+y2)·3by
=
3
2
b3y2-2b2y2+3by3

(4)(4b3y-2b2y2+12by3)÷(2by),
=2b2-3by+2y2

(z)(b+2y)2-(b+y)(b-y),
=b2+4by+4y2-b2+y2
=4by+zy2
当b=2,y=
1
2
时,原式=4×2×
1
2
+z×
1
4
=
21
4
考点梳理
整式的混合运算;整式的混合运算—化简求值.
(1)根据整数指数幂的运算性质计算;
(2)去括号,合并同类项即可;
(3)(4)根据整式的运算法则去括号即可;
(5)根据平方差和完全平方公式化简后,代入x,y的值计算.
本题考查了单项式的乘法,合并同类项,单项式乘多项式,多项式除单项式,完全平方公式,平方差公式,熟记整数指数幂的运算性质,平方差和完全平方公式是解决本题的关键.
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