1. 简便运算：
$48^{2}+48×24+12^{2}$；
$53.5^{2}×4 - 46.5^{2}×4$。
$48^{2}+48×24+12^{2}$；
$53.5^{2}×4 - 46.5^{2}×4$。
答案：1. (1)因为$(48+12)^{2}=48^{2}+48×24+12^{2}$，所以原式$=(48+12)^{2}=3600$。
(2)因为$(53.5+46.5)(53.5-46.5)=53.5^{2}-46.5^{2}$，所以原式$=4×(53.5^{2}-46.5^{2})=4×(53.5+46.5)(53.5-46.5)=4×100×7=2800$。
(2)因为$(53.5+46.5)(53.5-46.5)=53.5^{2}-46.5^{2}$，所以原式$=4×(53.5^{2}-46.5^{2})=4×(53.5+46.5)(53.5-46.5)=4×100×7=2800$。
2. 简便运算：
$1.03×0.97$；
$101×99 - 99.5^{2}$；
$101^{2}+99^{2}$。
$1.03×0.97$；
$101×99 - 99.5^{2}$；
$101^{2}+99^{2}$。
答案：2. (1)原式$=(1+0.03)×(1-0.03)=1-0.03^{2}=0.9991$。
(2)原式$=(100+1)×(100-1)-(100-\frac {1}{2})^{2}$
$=100^{2}-1^{2}-(100^{2}-100+\frac {1}{4})$
$=100^{2}-1-100^{2}+100-\frac {1}{4}=98\frac {3}{4}$。
(3)原式$=(100+1)^{2}+(100-1)^{2}$
$=100^{2}+2×1×100+1^{2}+100^{2}-2×1×100+1^{2}$
$=10000+200+1+10000-200+1$
$=20002$。
(2)原式$=(100+1)×(100-1)-(100-\frac {1}{2})^{2}$
$=100^{2}-1^{2}-(100^{2}-100+\frac {1}{4})$
$=100^{2}-1-100^{2}+100-\frac {1}{4}=98\frac {3}{4}$。
(3)原式$=(100+1)^{2}+(100-1)^{2}$
$=100^{2}+2×1×100+1^{2}+100^{2}-2×1×100+1^{2}$
$=10000+200+1+10000-200+1$
$=20002$。
3. 计算：
$(x + 2y + 1)(x - 2y + 1) - (x - 2y + 1)^{2}$；
$(a + b - c + d)(a - b + c + d)$。
$(x + 2y + 1)(x - 2y + 1) - (x - 2y + 1)^{2}$；
$(a + b - c + d)(a - b + c + d)$。
答案：3. (1)原式$=(x+1)^{2}-4y^{2}-(x-2y+1)^{2}$
$=(x+1)^{2}-4y^{2}-[(x+1)^{2}-4y(x+1)+4y^{2}]$
$=-8y^{2}+4xy+4y$。
(2)原式$=[(a+d)+(b-c)][(a+d)-(b-c)]$
$=(a+d)^{2}-(b-c)^{2}$
$=a^{2}+2ad+d^{2}-b^{2}+2bc-c^{2}$。
$=(x+1)^{2}-4y^{2}-[(x+1)^{2}-4y(x+1)+4y^{2}]$
$=-8y^{2}+4xy+4y$。
(2)原式$=[(a+d)+(b-c)][(a+d)-(b-c)]$
$=(a+d)^{2}-(b-c)^{2}$
$=a^{2}+2ad+d^{2}-b^{2}+2bc-c^{2}$。
4. 计算：$\frac{20252024^{2}}{20252023^{2}+20252025^{2}-2}$。
答案：4. 令$20252024=a$，则原式$=\frac {a^{2}}{(a-1)^{2}+(a+1)^{2}-2}=\frac {a^{2}}{2a^{2}+2-2}=\frac {a^{2}}{2a^{2}}=\frac {1}{2}$。
5. (1) 计算：$(5 + 1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}$；
(2) 计算：$\frac{1}{2}×(\frac{1}{2}+1)×(\frac{1}{2^{2}}+1)×(\frac{1}{2^{4}}+1)×(\frac{1}{2^{8}}+1)+1$。
(2) 计算：$\frac{1}{2}×(\frac{1}{2}+1)×(\frac{1}{2^{2}}+1)×(\frac{1}{2^{4}}+1)×(\frac{1}{2^{8}}+1)+1$。
答案：(1)
$\begin{aligned}&(5 + 1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5 - 1)(5 + 1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{2}-1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{4}-1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{8}-1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{16}-1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{32}-1)+\frac{1}{4}\\=&\frac{5^{32}}{4}-\frac{1}{4}+\frac{1}{4}\\=&\frac{5^{32}}{4}\end{aligned}$
(2)
$\begin{aligned}&\frac{1}{2}×(\frac{1}{2}+1)×(\frac{1}{2^{2}}+1)×(\frac{1}{2^{4}}+1)×(\frac{1}{2^{8}}+1)+1\\=&(1 - \frac{1}{2})(1 + \frac{1}{2})(1 + \frac{1}{2^{2}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{2}})(1 + \frac{1}{2^{2}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{4}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{8}})(1 + \frac{1}{2^{8}})+1\\=&1 - \frac{1}{2^{16}} + 1\\=&2 - \frac{1}{2^{16}}\\=&\frac{2^{17} - 1}{2^{16}}\end{aligned}$
$\begin{aligned}&(5 + 1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5 - 1)(5 + 1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{2}-1)(5^{2}+1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{4}-1)(5^{4}+1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{8}-1)(5^{8}+1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{16}-1)(5^{16}+1)+\frac{1}{4}\\=&\frac{1}{4}(5^{32}-1)+\frac{1}{4}\\=&\frac{5^{32}}{4}-\frac{1}{4}+\frac{1}{4}\\=&\frac{5^{32}}{4}\end{aligned}$
(2)
$\begin{aligned}&\frac{1}{2}×(\frac{1}{2}+1)×(\frac{1}{2^{2}}+1)×(\frac{1}{2^{4}}+1)×(\frac{1}{2^{8}}+1)+1\\=&(1 - \frac{1}{2})(1 + \frac{1}{2})(1 + \frac{1}{2^{2}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{2}})(1 + \frac{1}{2^{2}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{4}})(1 + \frac{1}{2^{4}})(1 + \frac{1}{2^{8}})+1\\=&(1 - \frac{1}{2^{8}})(1 + \frac{1}{2^{8}})+1\\=&1 - \frac{1}{2^{16}} + 1\\=&2 - \frac{1}{2^{16}}\\=&\frac{2^{17} - 1}{2^{16}}\end{aligned}$