1. $\sqrt{3}$ 的倒数是()

A.$-\sqrt{3}$
B.$-\dfrac{\sqrt{3}}{3}$
C.$-3$
D.$\dfrac{\sqrt{3}}{3}$

D

设$\sqrt{3}$的倒数为$x$，则$\sqrt{3} \cdot x = 1$，解得$x = \dfrac{1}{\sqrt{3}}$。对分母有理化，分子分母同乘以$\sqrt{3}$，得$x = \dfrac{\sqrt{3}}{3}$。

2. 化简 $\dfrac{-3\sqrt{3}}{\sqrt{18}}$ 的结果是()
A.$-\dfrac{\sqrt{6}}{2}$
B.$-\dfrac{\sqrt{3}}{2}$
C.$-\dfrac{3}{\sqrt{2}}$
D.$-\sqrt{3}$

A

原式为$\dfrac{-3\sqrt{3}}{\sqrt{18}}$，首先化简分母：$\sqrt{18} = \sqrt{9 × 2} = 3\sqrt{2}$，
代入后分式变为：$\dfrac{-3\sqrt{3}}{3\sqrt{2}} = -\dfrac{\sqrt{3}}{\sqrt{2}}$，
对分子分母同乘$\sqrt{2}$进行有理化，得：$-\dfrac{\sqrt{3} × \sqrt{2}}{\sqrt{2} × \sqrt{2}} = -\dfrac{\sqrt{6}}{2}$。
故选A。

3. 与 $2-\sqrt{3}$ 相乘,结果是 1 的数为()
A.$\sqrt{3}$
B.$2-\sqrt{3}$
C.$-2+\sqrt{3}$
D.$2+\sqrt{3}$

D

设所求数为 $x$，根据题意有 $x × (2 - \sqrt{3}) = 1$，
解得 $x = \frac{1}{2 - \sqrt{3}}$。
对分母有理化，得 $x = \frac{1 × (2 + \sqrt{3})}{(2 - \sqrt{3}) × (2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$。

4. 化简 $\dfrac{\sqrt{24}-\sqrt{54}}{\sqrt{3}}$ 的结果是.

-√2

$\begin{aligned}\frac{\sqrt{24}-\sqrt{54}}{\sqrt{3}}&=\frac{\sqrt{4×6}-\sqrt{9×6}}{\sqrt{3}}\\&=\frac{2\sqrt{6}-3\sqrt{6}}{\sqrt{3}}\\&=\frac{-\sqrt{6}}{\sqrt{3}}\\&=-\sqrt{\frac{6}{3}}\\&=-\sqrt{2}\end{aligned}$

5. $\sqrt{2}-1$ 的倒数为.

$\sqrt{2}+1$

根据倒数的定义，若两个数的乘积为$1$，则它们互为倒数，设$\sqrt{2}-1$的倒数为$x$，则$(\sqrt{2}-1)× x = 1$，所以$x=\frac{1}{\sqrt{2}-1}$。
对$\frac{1}{\sqrt{2}-1}$进行分母有理化，给分子分母同时乘以$\sqrt{2}+1$，得到$\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}$。
根据平方差公式$(a - b)(a + b)=a^2 - b^2$，这里$a = \sqrt{2}$，$b = 1$，则$(\sqrt{2}-1)(\sqrt{2}+1)=(\sqrt{2})^2 - 1^2=2 - 1 = 1$。
所以$\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1$。

6. 化简：
(1) $\dfrac{1}{3\sqrt{2}}$；
(2) $\dfrac{\sqrt{2}}{\sqrt{12}}$；
(3) $\dfrac{\sqrt{10}}{2\sqrt{5}}$。

(1)
$\begin{aligned}\dfrac{1}{3\sqrt{2}} \\= \dfrac{1 × \sqrt{2}}{3\sqrt{2} × \sqrt{2}} \\= \dfrac{\sqrt{2}}{6} \end{aligned}$
(2)
$\begin{aligned}\dfrac{\sqrt{2}}{\sqrt{12}} \\= \dfrac{\sqrt{2}}{2\sqrt{3}} \\= \dfrac{\sqrt{2} × \sqrt{3}}{2\sqrt{3} × \sqrt{3}} \\= \dfrac{\sqrt{6}}{6} \end{aligned}$
(3)
$\begin{aligned}\dfrac{\sqrt{10}}{2\sqrt{5}} \\= \dfrac{\sqrt{10} × \sqrt{5}}{2\sqrt{5} × \sqrt{5}} \\= \dfrac{\sqrt{50}}{10} \\= \dfrac{5\sqrt{2}}{10} \\= \dfrac{\sqrt{2}}{2}\end{aligned}$

7. 化简：
(1) $\dfrac{2}{\sqrt{3}-\sqrt{2}}$； (2) $\dfrac{\sqrt{2}}{\sqrt{5}+1}$；
(3) $\dfrac{1}{3-2\sqrt{2}}$； (4) $\dfrac{2}{1-\sqrt{2}}$；
(5) $\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1}$； (6) $\dfrac{1-\sqrt{3}}{2+\sqrt{3}}-\dfrac{4}{1+\sqrt{3}}$。

(1)
$\begin{aligned} \dfrac{2}{\sqrt{3}-\sqrt{2}} &= \dfrac{2(\sqrt{3} + \sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3} + \sqrt{2})} \\ &= \dfrac{2(\sqrt{3} + \sqrt{2})}{3 - 2} \\ &= 2\sqrt{3} + 2\sqrt{2} \end{aligned}$
(2)
$\begin{aligned} \dfrac{\sqrt{2}}{\sqrt{5}+1} &= \dfrac{\sqrt{2}(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)} \\ &= \dfrac{\sqrt{10}-\sqrt{2}}{5 - 1} \\ &= \dfrac{\sqrt{10}-\sqrt{2}}{4} \end{aligned}$
(3)
$\begin{aligned} \dfrac{1}{3-2\sqrt{2}} &= \dfrac{3 + 2\sqrt{2}}{(3-2\sqrt{2})(3 + 2\sqrt{2})} \\ &= \dfrac{3 + 2\sqrt{2}}{9 - 8} \\ &= 3 + 2\sqrt{2} \end{aligned}$
(4)
$\begin{aligned} \dfrac{2}{1-\sqrt{2}} &= \dfrac{2(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})} \\ &= \dfrac{2(1+\sqrt{2})}{1 - 2} \\ &= -2 - 2\sqrt{2} \end{aligned}$
(5)
$\begin{aligned} \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1} &= \dfrac{(\sqrt{6}+\sqrt{2})(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ &= \dfrac{3\sqrt{2}-\sqrt{6}+\sqrt{6}-\sqrt{2}}{3 - 1} \\ &= \sqrt{2} \end{aligned}$
(6)
$\begin{aligned} &\dfrac{1-\sqrt{3}}{2+\sqrt{3}}-\dfrac{4}{1+\sqrt{3}} \\ =& \dfrac{(1-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}-\dfrac{4(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} \\ =& \dfrac{2-\sqrt{3}-2\sqrt{3}+3}{4 - 3}-\dfrac{4 - 4\sqrt{3}}{1 - 3} \\ =& \dfrac{5 - 3\sqrt{3}}{1}-\dfrac{4 - 4\sqrt{3}}{-2} \\ =& 5 - 3\sqrt{3}+2 - 2\sqrt{3} \\ =& 7 - 5\sqrt{3} \end{aligned}$