试题

题目:
利用乘法公式计算:
(1)(2x-3y)2-(y+3x)(3x-y);
(2)(x+y)(x2+y2)(x-y)(x4+y4);
(3)(a-2b+3)(a+2b-3);
(4)[(x-y)2+(x+y)2](x2-y2);
(5)(m-n-3)2
答案
解:(1)原式=(2x-3y)2-(9x2-y2),
=(4x2+9y2-12xy)-9x2+y2
=10y2-12xy-5x2

(2)原式=(x+y)(x2+y2)(x-y)(x4+y4),
=(x2-y2)( x2+y2)(x4+y4),
=(x4-y4)(x4+y4),
=x8-y8

(3)原式=[a-(2b-3)][a+(2b-3)],
=a2-(2b-3)2
=a2-4b2-9+12b;

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

(5)原式=(m-n-3)(m-n-3),
=m2-mn-3m-mn+n2+3n-3m+3n+9,
=n2+m2-2mn-6m+6n+9.
解:(1)原式=(2x-3y)2-(9x2-y2),
=(4x2+9y2-12xy)-9x2+y2
=10y2-12xy-5x2

(2)原式=(x+y)(x2+y2)(x-y)(x4+y4),
=(x2-y2)( x2+y2)(x4+y4),
=(x4-y4)(x4+y4),
=x8-y8

(3)原式=[a-(2b-3)][a+(2b-3)],
=a2-(2b-3)2
=a2-4b2-9+12b;

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

(5)原式=(m-n-3)(m-n-3),
=m2-mn-3m-mn+n2+3n-3m+3n+9,
=n2+m2-2mn-6m+6n+9.
考点梳理
平方差公式;完全平方公式.
用完全平方公式和平方差公式结合合并同类项计算.
本题组考查了完全平方公式和平方差公式的灵活运用,计算时要认真仔细.
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