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12.若$x^2 - y^2 = -1$,则$(x - y)^{2025} · (x + y)^{2025} = (\ )$
A.1
B.-1
C.2
D.-2
A.1
B.-1
C.2
D.-2
答案
12. B
$(x - y)^{2025} · (x + y)^{2025} = [(x + y)(x - y)]^{2025}$(积的乘方的逆用)$= (x^2 - y^2)^{2025} = (-1)^{2025} = -1$。
$(x - y)^{2025} · (x + y)^{2025} = [(x + y)(x - y)]^{2025}$(积的乘方的逆用)$= (x^2 - y^2)^{2025} = (-1)^{2025} = -1$。
13. [2026临沂期末]若$x - y = 5$，则$x^2 - y^2 - 10y$的值是（
A.10
B.15
C.20
D.25
D
）A.10
B.15
C.20
D.25
答案
13. D
$\because x - y = 5, \therefore x^2 - y^2 - 10y = (x^2 - y^2) - 10y = (x + y)(x - y) - 10y = 5(x + y) - 10y = 5x + 5y - 10y = 5x - 5y = 5(x - y) = 5 × 5 = 25$。
$\because x - y = 5, \therefore x^2 - y^2 - 10y = (x^2 - y^2) - 10y = (x + y)(x - y) - 10y = 5(x + y) - 10y = 5x + 5y - 10y = 5x - 5y = 5(x - y) = 5 × 5 = 25$。
14. [2025 随州模拟]若 $ M = 2026^2 $，$ N = 2025 × 2027 $，则 $ M $
＞
$ N $.（填“＞”“＜”或“=”）答案
14. ＞
$\because M = 2026^2, N = 2025 × 2027 = (2026 - 1) × (2026 + 1) = 2026^2 - 1, \therefore M ＞ N$。
$\because M = 2026^2, N = 2025 × 2027 = (2026 - 1) × (2026 + 1) = 2026^2 - 1, \therefore M ＞ N$。
15. 如图，学校劳动实践基地有两块边长分别为a,b的正方形秧田A,B，其中不能使用的面积为M.
(1)用含a,M的代数式表示A中能使用的面积：
(2)若$a+b=10,a-b=5$,则A比B多出的使用面积为
(1)用含a,M的代数式表示A中能使用的面积：
$a^2 - M$
.(2)若$a+b=10,a-b=5$,则A比B多出的使用面积为
50
.答案
15. (1)$a^2 - M$；(2)50
A比B多出的使用面积为$(a^2 - M) - (b^2 - M) = a^2 - b^2 = (a + b)(a - b) = 10 × 5 = 50$。
A比B多出的使用面积为$(a^2 - M) - (b^2 - M) = a^2 - b^2 = (a + b)(a - b) = 10 × 5 = 50$。
16. 新趋势·代数推理 对于任意的正整数$n$，试说明：$(3n+1)(3n-1)-(3-n)(3+n)$能被10整除.
答案
16. 解:$(3n + 1)(3n - 1) - (3 - n)(3 + n) = (3n)^2 - 1 - (3^2 - n^2) = 9n^2 - 1 - 9 + n^2 = 10n^2 - 10 = 10(n^2 - 1)$。
$\because n$为正整数,$\therefore 10(n^2 - 1)$能被10整除,
$\therefore$ 对于任意正整数$n$,$(3n + 1)(3n - 1) - (3 - n)(3 + n)$能被10整除。
$\because n$为正整数,$\therefore 10(n^2 - 1)$能被10整除,
$\therefore$ 对于任意正整数$n$,$(3n + 1)(3n - 1) - (3 - n)(3 + n)$能被10整除。
17. 运算能力 小明遇到下面一个问题:计算$(2 + 1) × (2^2 + 1) × (2^4 + 1)$. 经观察,小明发现先将原式变形,可得特殊结构,再用平方差公式解决问题,具体解法如下:
$(2 + 1) × (2^2 + 1) × (2^4 + 1)$
$= (2 - 1) × (2 + 1) × (2^2 + 1) × (2^4 + 1)$
$= (2^2 - 1) × (2^2 + 1) × (2^4 + 1)$
$= (2^4 - 1) × (2^4 + 1) = 2^8 - 1$.
根据小明的方法,计算:
(1) $(1 - \frac{1}{2^2}) × (1 - \frac{1}{3^2}) × (1 - \frac{1}{4^2}) × … × (1 - \frac{1}{2025^2}) × (1 - \frac{1}{2026^2})$;
(2) $(1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2^{15}}$.
$(2 + 1) × (2^2 + 1) × (2^4 + 1)$
$= (2 - 1) × (2 + 1) × (2^2 + 1) × (2^4 + 1)$
$= (2^2 - 1) × (2^2 + 1) × (2^4 + 1)$
$= (2^4 - 1) × (2^4 + 1) = 2^8 - 1$.
根据小明的方法,计算:
(1) $(1 - \frac{1}{2^2}) × (1 - \frac{1}{3^2}) × (1 - \frac{1}{4^2}) × … × (1 - \frac{1}{2025^2}) × (1 - \frac{1}{2026^2})$;
(2) $(1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2^{15}}$.
答案
17. 解:
(1)原式
$=(1 - \frac{1}{2}) × (1 + \frac{1}{2}) × (1 - \frac{1}{3}) × (1 + \frac{1}{3}) × (1 - \frac{1}{4}) × (1 + \frac{1}{4}) × … × (1 - \frac{1}{2025}) × (1 + \frac{1}{2025}) × (1 - \frac{1}{2026}) × (1 + \frac{1}{2026})$
$= \frac{1}{2} × \frac{3}{2} × \frac{2}{3} × \frac{4}{3} × \frac{3}{4} × \frac{5}{4} × … × \frac{2024}{2025} × \frac{2026}{2025} × \frac{2025}{2026} × \frac{2027}{2026}$
$= \frac{1}{2} × \frac{2027}{2026} = \frac{2027}{4052}$。
(2)原式
$= 2 × (1 - \frac{1}{2})$(将式子恒等变形)$× [(1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2}) × (1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + (1 - \frac{1}{2}) × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^4}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^8}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^{16}}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × (1 - \frac{1}{2^{16}} + \frac{1}{2^{16}}) = 2$。
(1)原式
$=(1 - \frac{1}{2}) × (1 + \frac{1}{2}) × (1 - \frac{1}{3}) × (1 + \frac{1}{3}) × (1 - \frac{1}{4}) × (1 + \frac{1}{4}) × … × (1 - \frac{1}{2025}) × (1 + \frac{1}{2025}) × (1 - \frac{1}{2026}) × (1 + \frac{1}{2026})$
$= \frac{1}{2} × \frac{3}{2} × \frac{2}{3} × \frac{4}{3} × \frac{3}{4} × \frac{5}{4} × … × \frac{2024}{2025} × \frac{2026}{2025} × \frac{2025}{2026} × \frac{2027}{2026}$
$= \frac{1}{2} × \frac{2027}{2026} = \frac{2027}{4052}$。
(2)原式
$= 2 × (1 - \frac{1}{2})$(将式子恒等变形)$× [(1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2}) × (1 + \frac{1}{2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + (1 - \frac{1}{2}) × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^2}) × (1 + \frac{1}{2^2}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^4}) × (1 + \frac{1}{2^4}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^8}) × (1 + \frac{1}{2^8}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × [(1 - \frac{1}{2^{16}}) + \frac{1}{2} × \frac{1}{2^{15}}]$
$= 2 × (1 - \frac{1}{2^{16}} + \frac{1}{2^{16}}) = 2$。
