登录
三、解答题
5. 化简：
（1）$\dfrac{1}{\sqrt{5}}$；
（2）$\sqrt{2\dfrac{1}{2}}$；
（3）$\dfrac{\sqrt{3}x}{\sqrt{6x}}(x＞0)$；
（4）$\dfrac{\sqrt{18m}}{\sqrt{2n}}(m≥0,n＞0)$.
5. 化简：
（1）$\dfrac{1}{\sqrt{5}}$；
（2）$\sqrt{2\dfrac{1}{2}}$；
（3）$\dfrac{\sqrt{3}x}{\sqrt{6x}}(x＞0)$；
（4）$\dfrac{\sqrt{18m}}{\sqrt{2n}}(m≥0,n＞0)$.
答案
解：
（1）$\dfrac{1}{\sqrt{5}}=\dfrac{1×\sqrt{5}}{\sqrt{5}×\sqrt{5}}=\dfrac{\sqrt{5}}{5}$；
（2）$\sqrt{2\dfrac{1}{2}}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}×\sqrt{2}}{\sqrt{2}×\sqrt{2}}=\dfrac{\sqrt{10}}{2}$；
（3）$\dfrac{\sqrt{3}x}{\sqrt{6x}}=\sqrt{\dfrac{3x^2}{6x}}=\sqrt{\dfrac{x}{2}}=\dfrac{\sqrt{x}×\sqrt{2}}{\sqrt{2}×\sqrt{2}}=\dfrac{\sqrt{2x}}{2}$；
（4）$\dfrac{\sqrt{18m}}{\sqrt{2n}}=\sqrt{\dfrac{18m}{2n}}=\sqrt{\dfrac{9m}{n}}=\dfrac{3\sqrt{m}}{\sqrt{n}}=\dfrac{3\sqrt{m}×\sqrt{n}}{\sqrt{n}×\sqrt{n}}=\dfrac{3\sqrt{mn}}{n}$。
（1）$\dfrac{1}{\sqrt{5}}=\dfrac{1×\sqrt{5}}{\sqrt{5}×\sqrt{5}}=\dfrac{\sqrt{5}}{5}$；
（2）$\sqrt{2\dfrac{1}{2}}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}×\sqrt{2}}{\sqrt{2}×\sqrt{2}}=\dfrac{\sqrt{10}}{2}$；
（3）$\dfrac{\sqrt{3}x}{\sqrt{6x}}=\sqrt{\dfrac{3x^2}{6x}}=\sqrt{\dfrac{x}{2}}=\dfrac{\sqrt{x}×\sqrt{2}}{\sqrt{2}×\sqrt{2}}=\dfrac{\sqrt{2x}}{2}$；
（4）$\dfrac{\sqrt{18m}}{\sqrt{2n}}=\sqrt{\dfrac{18m}{2n}}=\sqrt{\dfrac{9m}{n}}=\dfrac{3\sqrt{m}}{\sqrt{n}}=\dfrac{3\sqrt{m}×\sqrt{n}}{\sqrt{n}×\sqrt{n}}=\dfrac{3\sqrt{mn}}{n}$。
6. 计算：
（1）$\sqrt{12x}÷\dfrac{3}{7}\sqrt{y}(x≥0,y＞0)$；
（2）$x\sqrt{xy}÷ y\sqrt{\dfrac{x}{y}}·\sqrt{\dfrac{y}{x}}(x＞0,y＞0)$.
（1）$\sqrt{12x}÷\dfrac{3}{7}\sqrt{y}(x≥0,y＞0)$；
（2）$x\sqrt{xy}÷ y\sqrt{\dfrac{x}{y}}·\sqrt{\dfrac{y}{x}}(x＞0,y＞0)$.
答案
解：
（1）$\sqrt{12x}÷\dfrac{3}{7}\sqrt{y}$
$=(1÷\dfrac{3}{7})×(\sqrt{12x}÷\sqrt{y})$
$=\dfrac{7}{3}×\sqrt{\dfrac{12x}{y}}$
$=\dfrac{7}{3}×\sqrt{\dfrac{4×3xy}{y^2}}$
$=\dfrac{7}{3}×\dfrac{2\sqrt{3xy}}{y}$
$=\dfrac{14\sqrt{3xy}}{3y}$
（2）$x\sqrt{xy}÷ y\sqrt{\dfrac{x}{y}}·\sqrt{\dfrac{y}{x}}$
$=\dfrac{x}{y}×(\sqrt{xy}÷\sqrt{\dfrac{x}{y}}×\sqrt{\dfrac{y}{x}})$
$=\dfrac{x}{y}×\sqrt{xy÷\dfrac{x}{y}×\dfrac{y}{x}}$
$=\dfrac{x}{y}×\sqrt{xy×\dfrac{y}{x}×\dfrac{y}{x}}$
$=\dfrac{x}{y}×\sqrt{\dfrac{y^3}{x}}$
$=\dfrac{x}{y}×\dfrac{y\sqrt{xy}}{x}$
$=\sqrt{xy}$
（1）$\sqrt{12x}÷\dfrac{3}{7}\sqrt{y}$
$=(1÷\dfrac{3}{7})×(\sqrt{12x}÷\sqrt{y})$
$=\dfrac{7}{3}×\sqrt{\dfrac{12x}{y}}$
$=\dfrac{7}{3}×\sqrt{\dfrac{4×3xy}{y^2}}$
$=\dfrac{7}{3}×\dfrac{2\sqrt{3xy}}{y}$
$=\dfrac{14\sqrt{3xy}}{3y}$
（2）$x\sqrt{xy}÷ y\sqrt{\dfrac{x}{y}}·\sqrt{\dfrac{y}{x}}$
$=\dfrac{x}{y}×(\sqrt{xy}÷\sqrt{\dfrac{x}{y}}×\sqrt{\dfrac{y}{x}})$
$=\dfrac{x}{y}×\sqrt{xy÷\dfrac{x}{y}×\dfrac{y}{x}}$
$=\dfrac{x}{y}×\sqrt{xy×\dfrac{y}{x}×\dfrac{y}{x}}$
$=\dfrac{x}{y}×\sqrt{\dfrac{y^3}{x}}$
$=\dfrac{x}{y}×\dfrac{y\sqrt{xy}}{x}$
$=\sqrt{xy}$
7. 在表格中填数，使每一行、每一列、每条对角线上的 3 个数的乘积都是 1.
答案
解：设表格各位置的数如下：
$\begin{array}{|c|c|c|}\hlinea & b & \sqrt{2} \\\hlinec & 1 & d \\\hlinee & f & \sqrt{3} \\\hline\end{array}$
1. 由主对角线乘积为1：$a × 1 × \sqrt{3} = 1$，得$a = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$。
2. 由第一行乘积为1：$a × b × \sqrt{2} = 1$，代入$a=\frac{\sqrt{3}}{3}$，得$b = \frac{1}{a × \sqrt{2}} = \frac{1}{\frac{\sqrt{3}}{3} × \sqrt{2}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$。
3. 由副对角线乘积为1：$\sqrt{2} × 1 × e = 1$，得$e = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$。
4. 由第一列乘积为1：$a × c × e = 1$，代入$a=\frac{\sqrt{3}}{3}$，$e=\frac{\sqrt{2}}{2}$，得$c = \frac{1}{a × e} = \frac{1}{\frac{\sqrt{3}}{3} × \frac{\sqrt{2}}{2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$。
5. 由第二行乘积为1：$c × 1 × d = 1$，代入$c=\sqrt{6}$，得$d = \frac{1}{c} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$。
6. 由第三行乘积为1：$e × f × \sqrt{3} = 1$，代入$e=\frac{\sqrt{2}}{2}$，得$f = \frac{1}{e × \sqrt{3}} = \frac{1}{\frac{\sqrt{2}}{2} × \sqrt{3}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$。
最终填数后的表格为：
$\begin{array}{|c|c|c|}\hline\frac{\sqrt{3}}{3} & \frac{\sqrt{6}}{2} & \sqrt{2} \\\hline\sqrt{6} & 1 & \frac{\sqrt{6}}{6} \\\hline\frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{3} & \sqrt{3} \\\hline\end{array}$
$\begin{array}{|c|c|c|}\hlinea & b & \sqrt{2} \\\hlinec & 1 & d \\\hlinee & f & \sqrt{3} \\\hline\end{array}$
1. 由主对角线乘积为1：$a × 1 × \sqrt{3} = 1$，得$a = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$。
2. 由第一行乘积为1：$a × b × \sqrt{2} = 1$，代入$a=\frac{\sqrt{3}}{3}$，得$b = \frac{1}{a × \sqrt{2}} = \frac{1}{\frac{\sqrt{3}}{3} × \sqrt{2}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$。
3. 由副对角线乘积为1：$\sqrt{2} × 1 × e = 1$，得$e = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$。
4. 由第一列乘积为1：$a × c × e = 1$，代入$a=\frac{\sqrt{3}}{3}$，$e=\frac{\sqrt{2}}{2}$，得$c = \frac{1}{a × e} = \frac{1}{\frac{\sqrt{3}}{3} × \frac{\sqrt{2}}{2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$。
5. 由第二行乘积为1：$c × 1 × d = 1$，代入$c=\sqrt{6}$，得$d = \frac{1}{c} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$。
6. 由第三行乘积为1：$e × f × \sqrt{3} = 1$，代入$e=\frac{\sqrt{2}}{2}$，得$f = \frac{1}{e × \sqrt{3}} = \frac{1}{\frac{\sqrt{2}}{2} × \sqrt{3}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$。
最终填数后的表格为：
$\begin{array}{|c|c|c|}\hline\frac{\sqrt{3}}{3} & \frac{\sqrt{6}}{2} & \sqrt{2} \\\hline\sqrt{6} & 1 & \frac{\sqrt{6}}{6} \\\hline\frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{3} & \sqrt{3} \\\hline\end{array}$
8. 阅读下列材料，然后回答问题.
在进行二次根式的化简与运算时，我们有时会碰到形如$\dfrac{2}{\sqrt{3}+1}$的式子，我们还可以将其进一步化简：$\dfrac{2}{\sqrt{3}+1}=\dfrac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\sqrt{3}-1$.以上这种化简的步骤叫作分母有理化.
（1）化简：$\dfrac{2}{\sqrt{7}-\sqrt{5}}$.
（2）化简：$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+···+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n-1}}$.
在进行二次根式的化简与运算时，我们有时会碰到形如$\dfrac{2}{\sqrt{3}+1}$的式子，我们还可以将其进一步化简：$\dfrac{2}{\sqrt{3}+1}=\dfrac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\sqrt{3}-1$.以上这种化简的步骤叫作分母有理化.
（1）化简：$\dfrac{2}{\sqrt{7}-\sqrt{5}}$.
（2）化简：$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+···+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n-1}}$.
答案
解：（1）$\dfrac{2}{\sqrt{7}-\sqrt{5}}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{7-5}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{2}$
$=\sqrt{7}+\sqrt{5}$
（2）$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+···+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n-1}}$
$=\dfrac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}+\dfrac{\sqrt{5}-\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}+\dfrac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}+···+\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n+1}-\sqrt{2n-1})}$
$=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+···+\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{2}$
$=\dfrac{(\sqrt{3}-1)+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+···+(\sqrt{2n+1}-\sqrt{2n-1})}{2}$
$=\dfrac{\sqrt{2n+1}-1}{2}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{7-5}$
$=\dfrac{2(\sqrt{7}+\sqrt{5})}{2}$
$=\sqrt{7}+\sqrt{5}$
（2）$\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{5}+\sqrt{3}}+\dfrac{1}{\sqrt{7}+\sqrt{5}}+···+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n-1}}$
$=\dfrac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}+\dfrac{\sqrt{5}-\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}+\dfrac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}+···+\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n+1}-\sqrt{2n-1})}$
$=\dfrac{\sqrt{3}-1}{2}+\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+···+\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{2}$
$=\dfrac{(\sqrt{3}-1)+(\sqrt{5}-\sqrt{3})+(\sqrt{7}-\sqrt{5})+···+(\sqrt{2n+1}-\sqrt{2n-1})}{2}$
$=\dfrac{\sqrt{2n+1}-1}{2}$
