1. 計算:
(1)$(a-2b+1)(a+2b+1)$; (2)$(x+2y-1)^{2}$;
(3)$(x-1)^{2}(x+1)^{2}(x^{2}+1)^{2}$; (4)$(2x-1)^{2}-(x-2)^{2}$.
答案：(1)解：原式=[(a+1)-2b][(a+1)+2b]
=(a+1)2-(2b)2
=a2+2a+1-4b2
(2)解：原式=[(x+2y)-1]2
=(x+2y)2-2(x+2y)×1+12
=x2+4xy+4y2-2x-4y+1
(3)解：原式=[(x-1)(x+1)]2(x2+1)2
=(x2-1)2(x2+1)2
=[(x2-1)(x2+1)]2
=(x?-1)2
=x?-2x?+1
(4)解：原式=(4x2-4x+1)-(x2-4x+4)
=4x2-4x+1-x2+4x-4
=3x2-3
2. 計算:$(x+y+z)(x+y-z)-(x+y+z)^{2}$.
答案：解：原式$=[(x+y)^2 - z^2] - [(x+y)^2 + 2z(x+y) + z^2]$
$=(x+y)^2 - z^2 - (x+y)^2 - 2z(x+y) - z^2$
$=-2z^2 - 2xz - 2yz$
3. 計算:$(x-2y-z)(x+2y-z)-(x+z)^{2}$.
答案：解：原式$=[(x+y)^2 - z^2] - [(x+y)^2 + 2z(x+y) + z^2]$
$=(x+y)^2 - z^2 - (x+y)^2 - 2z(x+y) - z^2$
$=-2z^2 - 2xz - 2yz$
4. 計算:$(2^{4}-1)(2^{4}+1)(2^{8}+1)(2^{16}+1)(2^{32}+1)$.
答案：解:原式$=(2^{8}-1)(2^{8}+1)(2^{16}+1)(2^{32}+1)=(2^{16}-1)\cdot$$(2^{16}+1)(2^{32}+1)=(2^{32}-1)(2^{32}+1)=2^{64}-1.$
5. 已知代數(shù)式$(m-1)^{2}+(m+n)(m-n)+n^{2}$.
(1)化簡這個代數(shù)式;
(2)若$m^{2}-m-3= 0$,求這個代數(shù)式的值.
答案：解:(1)$(m-1)^{2}+(m+n)(m-n)+n^{2}$$=m^{2}-2m+1+m^{2}-n^{2}+n^{2}$$=2m^{2}-2m+1.$(2)$\because m^{2}-m-3=0,$$\therefore m^{2}-m=3,$$\therefore 2m^{2}-2m+1=2(m^{2}-m)+1=2×3+1=7.$
