试题

题目:
计算
(1)(x+y)2(x-y)2-(x-y)(x+y)(x2+y2);
(2)(2x-3)(x-2)-2(x-1)2
(3)[(
1
2
x-y)2+(
1
2
x+y)2](
1
2
x2-2y2)

(4)(a+1)(a+2)(a+3)(a+4).
答案
解:(1)原式=(x2-y22-(x2-y2)(x2+y2),
=(x2-y2)(x2-y2-x2-y2),
=-2x2y2+2y4

(2)原式=(2x-4+1)(x-2)-2(x-1)2
=2(x-2)2-2(x-1)2+x-2,
=2(x-2+x-1)(x-2-x+1)+x-2,
=2(2x-3)(-1)+x-2,
=-3x+4;

(3)原式=[(
1
2
x-y)2+2(
1
2
x-y)(
1
2
x+y)+(
1
2
x+y)2-2(
1
2
x-y)(
1
2
x+y)](
1
2
x2-2y2),
=(
1
2
x2+2y2)(
1
2
x2-2y2),
=
1
4
x4+4y4

(4)原式=(a+1)(a+4)(a+2)(a+3);
=(a2+5a+4)(a2+5a+6),
=(a2+5a)2+10(a2+5a)+24,
=a4+10a3+35a2+50a+24.
解:(1)原式=(x2-y22-(x2-y2)(x2+y2),
=(x2-y2)(x2-y2-x2-y2),
=-2x2y2+2y4

(2)原式=(2x-4+1)(x-2)-2(x-1)2
=2(x-2)2-2(x-1)2+x-2,
=2(x-2+x-1)(x-2-x+1)+x-2,
=2(2x-3)(-1)+x-2,
=-3x+4;

(3)原式=[(
1
2
x-y)2+2(
1
2
x-y)(
1
2
x+y)+(
1
2
x+y)2-2(
1
2
x-y)(
1
2
x+y)](
1
2
x2-2y2),
=(
1
2
x2+2y2)(
1
2
x2-2y2),
=
1
4
x4+4y4

(4)原式=(a+1)(a+4)(a+2)(a+3);
=(a2+5a+4)(a2+5a+6),
=(a2+5a)2+10(a2+5a)+24,
=a4+10a3+35a2+50a+24.
考点梳理
整式的混合运算.
(1)本题可用平方差公式进行化简;
(2)(2x-3)可写成2(x-2)+1,然后根据提取公因式法和平方差公式进行化简;
(3)可先将大括号内的式子配成完全平方公式,然后再化简;
(4)第一项与第四项相乘,二与三相乘,再用分配律和乘法公式计算.
本题考查整式的综合运算能力,要注意对各公式的熟练应用.
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