答案
解:(1)原式=(x
2-y
2)
2-(x
2-y
2)(x
2+y
2),
=(x
2-y
2)(x
2-y
2-x
2-y
2),
=-2x
2y
2+2y
4;
(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(
x-y)(
x+y)+(
x+y)
2-2(
x-y)(
x+y)](
x
2-2y
2),
=(
x
2+2y
2)(
x
2-2y
2),
=
x
4+4y
4;
(4)原式=(a+1)(a+4)(a+2)(a+3);
=(a
2+5a+4)(a
2+5a+6),
=(a
2+5a)
2+10(a
2+5a)+24,
=a
4+10a
3+35a
2+50a+24.
解:(1)原式=(x
2-y
2)
2-(x
2-y
2)(x
2+y
2),
=(x
2-y
2)(x
2-y
2-x
2-y
2),
=-2x
2y
2+2y
4;
(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(
x-y)(
x+y)+(
x+y)
2-2(
x-y)(
x+y)](
x
2-2y
2),
=(
x
2+2y
2)(
x
2-2y
2),
=
x
4+4y
4;
(4)原式=(a+1)(a+4)(a+2)(a+3);
=(a
2+5a+4)(a
2+5a+6),
=(a
2+5a)
2+10(a
2+5a)+24,
=a
4+10a
3+35a
2+50a+24.