1. 把$\frac{\sqrt{3a}}{\sqrt{12ab}}$分母有理化后得 ( )
A. $4b$
B. $2\sqrt{b}$
C. $\frac{1}{2}\sqrt{b}$
D. $\frac{\sqrt{b}}{2b}$

D

2. 化简：
(1)$\frac{6}{\sqrt{3}}=$_______；(2)$\frac{-9\sqrt{2}}{\sqrt{27}}=$_______；

(1)$2\sqrt{3}$ (2)$-\sqrt{6}$

3. 满足$\frac{1}{\sqrt{3}-\sqrt{2}}＜x＜\frac{2}{\sqrt{6}-\sqrt{5}}$的整数$x$的个数是 ( )
A. $4$
B. $5$
C. $6$
D. $7$

C 解析：$\because \frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\approx 3.1$，$\frac{2}{\sqrt{6}-\sqrt{5}}=2\times(\sqrt{6}+\sqrt{5})\approx 9.4$，$\therefore$满足$\frac{1}{\sqrt{3}-\sqrt{2}}＜x＜\frac{2}{\sqrt{6}-\sqrt{5}}$的整数$x$有$4,5,6,7,8,9$，共6个. 故选C.

4. 一题多解 (2024·德州期末) 已知$m = \sqrt{6} + 2$，$n = \sqrt{6} - 2$，则$\frac{1}{m} + \frac{1}{n}$的值为_______.

$\sqrt{6}$ 解析：$\frac{1}{m}+\frac{1}{n}=\frac{1}{\sqrt{6}+2}+\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}-2+\sqrt{6}+2}{(\sqrt{6}+2)(\sqrt{6}-2)}=\frac{2\sqrt{6}}{2}=\sqrt{6}$.
一题多解
$\because m = \sqrt{6}+2$，$n=\sqrt{6}-2$，$\therefore m + n=\sqrt{6}+2+\sqrt{6}-2=2\sqrt{6}$，$mn=(\sqrt{6}+2)\cdot(\sqrt{6}-2)=2$，$\therefore \frac{1}{m}+\frac{1}{n}=\frac{m + n}{mn}=\frac{2\sqrt{6}}{2}=\sqrt{6}$.

5. 若$[x]$表示不超过$x$的最大整数，$A = \frac{1}{1 - \sqrt[4]{3}} + \frac{1}{1 + \sqrt[4]{3}} + (\frac{1}{1 - \sqrt[4]{3}})^0$，则$[A] =$_______.

-2 解析：$A=\frac{1}{1-\sqrt[4]{3}}+\frac{1}{1+\sqrt[4]{3}}+\left(\frac{1}{1-\sqrt[4]{3}}\right)^0=\frac{1+\sqrt[4]{3}}{(1-\sqrt[4]{3})(1+\sqrt[4]{3})}+\frac{1-\sqrt[4]{3}}{(1+\sqrt[4]{3})(1-\sqrt[4]{3})}+1$，那么$A=\frac{1+\sqrt[4]{3}}{1-\sqrt{3}}+\frac{1-\sqrt[4]{3}}{1-\sqrt{3}}+1=\frac{2}{1-\sqrt{3}}+1=\frac{2(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}+1=-1-\sqrt{3}+1=-\sqrt{3}$，$\therefore -2＜A＜-1$，$[A]=-2$.

6. 已知$m$是正整数，$a = \frac{\sqrt{m + 1} - \sqrt{m}}{\sqrt{m + 1} + \sqrt{m}}$，$b = \frac{\sqrt{m + 1} + \sqrt{m}}{\sqrt{m + 1} - \sqrt{m}}$，$a + b + 3ab = 2025$，求$m$的值.

$\because a=\frac{\sqrt{m + 1}-\sqrt{m}}{\sqrt{m + 1}+\sqrt{m}}$，$b=\frac{\sqrt{m + 1}+\sqrt{m}}{\sqrt{m + 1}-\sqrt{m}}$，分母有理化得$a = (\sqrt{m + 1}-\sqrt{m})^2$，$b = (\sqrt{m + 1}+\sqrt{m})^2$，$ab = 1$，$\therefore a + b = (\sqrt{m + 1}-\sqrt{m})^2+(\sqrt{m + 1}+\sqrt{m})^2=4m + 2$.$\because a + b+3ab = 2025$，$\therefore 4m + 2+3 = 2025$，$\therefore m = 505$.

7. 化简：(1)$\frac{x - y}{\sqrt{x} - \sqrt{y}}$；(2)$\frac{7 + 4\sqrt{3}}{2 + \sqrt{3}}$；(3)$\frac{x - 2 + \sqrt{x^2 - 4}}{x + 2 + \sqrt{x^2 - 4}}(x ＞ 2)$.

(1)$\frac{x - y}{\sqrt{x}-\sqrt{y}}=\frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}$.
(2)$\frac{7 + 4\sqrt{3}}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}$.
(3)$\because x＞2$，$\therefore \frac{x - 2+\sqrt{x^2 - 4}}{x + 2+\sqrt{x^2 - 4}}=\frac{(\sqrt{x - 2})^2+\sqrt{(x + 2)(x - 2)}}{(\sqrt{x + 2})^2+\sqrt{(x + 2)(x - 2)}}=\frac{\sqrt{x - 2}(\sqrt{x - 2}+\sqrt{x + 2})}{\sqrt{x + 2}(\sqrt{x + 2}+\sqrt{x - 2})}=\frac{\sqrt{x - 2}}{\sqrt{x + 2}}=\frac{\sqrt{x^2 - 4}}{x + 2}$.

8. 化简：$\frac{\sqrt{2} + 2\sqrt{3} + \sqrt{5}}{(\sqrt{2} + \sqrt{3})(\sqrt{3} + \sqrt{5})}$.

$\frac{\sqrt{2}+2\sqrt{3}+\sqrt{5}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}+\sqrt{5})}=\frac{(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{5})}{(\sqrt{2}+\sqrt{3})(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}+\sqrt{5})}+\frac{\sqrt{3}+\sqrt{5}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}+\sqrt{5})}=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{5}+\sqrt{3}-2\sqrt{2}}{2}$.

9. 化简：$\frac{\sqrt{2} - \sqrt{3} + \sqrt{5}}{\sqrt{2} + \sqrt{3} + \sqrt{5}}$.

原式$=\frac{(\sqrt{2}+\sqrt{5}-\sqrt{3})^2}{(\sqrt{2}+\sqrt{5}+\sqrt{3})(\sqrt{2}+\sqrt{5}-\sqrt{3})}=\frac{(\sqrt{2}+\sqrt{5})^2-3}{\sqrt{10}-\sqrt{6}-\sqrt{15}+5}=\frac{(\sqrt{10}-\sqrt{6}-\sqrt{15}+5)(\sqrt{10}-2)}{(\sqrt{10}+2)(\sqrt{10}-2)}=\frac{3\sqrt{10}-3\sqrt{6}}{6}=\frac{\sqrt{10}-\sqrt{6}}{2}$.