2026年学霸题中题八年级数学下册苏科版第164页解析答案

1. (2025·苏州期末)定义:我们将$(\sqrt{a}+\sqrt{b})$与$(\sqrt{a}-\sqrt{b})$称为一对“对偶式”.
因为$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=(\sqrt{a})^{2}-(\sqrt{b})^{2}=a - b$,可以有效地去掉根号,所以有一些题可以通过构造“对偶式”来解决.
例如:已知$\sqrt{12 - x}-\sqrt{8 - x}=2$,求$\sqrt{12 - x}+\sqrt{8 - x}$的值,可以这样解答:
因为$(\sqrt{12 - x}-\sqrt{8 - x})×(\sqrt{12 - x}+\sqrt{8 - x})=(\sqrt{12 - x})^{2}-(\sqrt{8 - x})^{2}=12 - x - 8 + x = 4$,
所以$\sqrt{12 - x}+\sqrt{8 - x}=2$.
根据以上材料,理解并运用材料提供的方法,解答下列问题:
(1)已知$\sqrt{18 - x}+\sqrt{6 - x}=6$,则$\sqrt{18 - x}-\sqrt{6 - x}=$

;
(2)化简:$\frac{1}{\sqrt{6}+\sqrt{5}}=$

;$\frac{1}{\sqrt{5}-\sqrt{3}}=$

;
(3)计算:$(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+···+\frac{1}{\sqrt{2023}+\sqrt{2025}})×(1+\sqrt{2025})$.

答案：(1)2　解析:因为$(\sqrt{18 - x} + \sqrt{6 - x})(\sqrt{18 - x} - \sqrt{6 - x}) = (\sqrt{18 - x})^2 - (\sqrt{6 - x})^2 = 18 - x - 6 + x = 12$，所以$\sqrt{18 - x} - \sqrt{6 - x} = 12 ÷ 6 = 2$。
(2)$\sqrt{6} - \sqrt{5}$ $\frac{\sqrt{5} + \sqrt{3}}{2}$ 解析:$\frac{1}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} - \sqrt{5}}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})} = \sqrt{6} - \sqrt{5}$，$\frac{1}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{5} + \sqrt{3}}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \frac{\sqrt{5} + \sqrt{3}}{2}$。
(3)原式$= [ \frac{\sqrt{3} - 1}{(\sqrt{3} + 1)(\sqrt{3} - 1)} + \frac{\sqrt{5} - \sqrt{3}}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})} + \frac{\sqrt{7} - \sqrt{5}}{(\sqrt{7} + \sqrt{5})(\sqrt{7} - \sqrt{5})} + ··· + \frac{\sqrt{2025} - \sqrt{2023}}{(\sqrt{2025} + \sqrt{2023})(\sqrt{2025} - \sqrt{2023})} ] × (\sqrt{2025} + 1) = ( \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + ··· + \frac{\sqrt{2025} - \sqrt{2023}}{2} ) × (\sqrt{2025} + 1) = \frac{1}{2}(\sqrt{3} - 1 + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + ··· + \sqrt{2025} - \sqrt{2023}) × (\sqrt{2025} + 1) = \frac{1}{2}(\sqrt{2025} - 1) × (\sqrt{2025} + 1) = \frac{1}{2} × (2025 - 1) = 1012$。

2. (2025·泰州月考)阅读材料:一般情形下等式$\frac{1}{x}+\frac{1}{y}=1$不成立,但有些特殊实数可以使它成立,例如:$x = 2,y = 2$时,$\frac{1}{2}+\frac{1}{2}=1$成立,我们称$(2,2)$是使$\frac{1}{x}+\frac{1}{y}=1$成立的“倒立数对”,请完成下列问题:
(1)数对$(\frac{4}{3},4),(1,1)$中,使$\frac{1}{x}+\frac{1}{y}=1$成立的“倒立数对”是

$(\frac{4}{3}, 4)$

;
(2)若$(5 - t,5 + t)$是使$\frac{1}{x}+\frac{1}{y}=1$成立的“倒立数对”,求$t$的值;
(3)若$(m,n)$是使$\frac{1}{x}+\frac{1}{y}=1$成立的“倒立数对”,且$a = b + m,b = c + n$,求代数式$\frac{12(a - b)(b - c)}{4(a - c)^{2}}$为整数时,$m + n$的值.

答案：(1)$(\frac{4}{3}, 4)$ 解析:将$(\frac{4}{3}, 4)$，$(1, 1)$分别代入$\frac{1}{x} + \frac{1}{y} = 1$，得到$\frac{1}{\frac{4}{3}} + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} = 1$，$\frac{1}{1} + \frac{1}{1} = 2 ≠ 1$，$\therefore$ 使$\frac{1}{x} + \frac{1}{y} = 1$成立的“倒立数对”是$(\frac{4}{3}, 4)$。
(2)$\because (5 - t, 5 + t)$是使$\frac{1}{x} + \frac{1}{y} = 1$成立的“倒立数对”，$\therefore \frac{1}{5 - t} + \frac{1}{5 + t} = 1$，解得$t = \pm \sqrt{15}$，经检验，$t = \pm \sqrt{15}$是$\frac{1}{5 - t} + \frac{1}{5 + t} = 1$的根，$\therefore t = \pm \sqrt{15}$。
(3)$\because (m, n)$是使$\frac{1}{x} + \frac{1}{y} = 1$成立的“倒立数对”，$\therefore \frac{1}{m} + \frac{1}{n} = 1$。$\because a = b + m$，$b = c + n$，$\therefore m = a - b$，$n = b - c$，$\therefore \frac{1}{a - b} + \frac{1}{b - c} = 1$，$m + n = a - b + b - c = a - c$，$\therefore b - c + a - b = (a - b)(b - c)$，$\therefore a - c = (a - b)(b - c)$，$\therefore \frac{12(a - b)(b - c)}{4(a - c)^2} = \frac{12(a - c)}{4(a - c)^2} = \frac{3}{a - c}$。$\because$ 代数式$\frac{12(a - b)(b - c)}{4(a - c)^2}$为整数，$\therefore a - c = \pm 3$或$a - c = \pm 1$，$\therefore m + n = \pm 3$或$m + n = \pm 1$。