问题
解答题
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)[(
(4)(a+1)(a+2)(a+3)(a+4). |
答案
(1)原式=(x2-y2)2-(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=[(
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 2
=
x4+4y4;1 4
(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.