新知梳理
1. 二次根式的除法法则：$\frac{\sqrt{a}}{\sqrt{b}} =$_________$(a \geq 0, b ＞ 0)$，两个二次根式相除可采用根号前系数对应相除，再乘根号内被开方数对应相除后的根式。
2. $\sqrt{\frac{a}{b}} =$_________$(a \geq 0, b ＞ 0)$，商式的算术平方根等于被除式的算术平方根除以除式的算术平方根。
1. 二次根式的除法法则：$\frac{\sqrt{a}}{\sqrt{b}} =$_________$(a \geq 0, b ＞ 0)$，两个二次根式相除可采用根号前系数对应相除，再乘根号内被开方数对应相除后的根式。
2. $\sqrt{\frac{a}{b}} =$_________$(a \geq 0, b ＞ 0)$，商式的算术平方根等于被除式的算术平方根除以除式的算术平方根。
答案：１. $\sqrt{\frac{a}{b}}$ ２. $\frac{\sqrt{a}}{\sqrt{b}}$
1. 计算$\sqrt{8} ÷ \sqrt{2}$的结果是(
A.$\sqrt{10}$
B.$\sqrt{6}$
C.4
D.2
D
)A.$\sqrt{10}$
B.$\sqrt{6}$
C.4
D.2
答案：１.D
解析：
$\sqrt{8} ÷ \sqrt{2} = \sqrt{8÷2} = \sqrt{4} = 2$，结果是D。
2. 计算$\sqrt{20} ÷ □ = \sqrt{5}$，则$□$中的数是(
A.4
B.$\sqrt{10}$
C.2
D.$\sqrt{2}$
C
)A.4
B.$\sqrt{10}$
C.2
D.$\sqrt{2}$
答案：２.C
解析：
设$□$中的数为$x$，则$\sqrt{20} ÷ x = \sqrt{5}$，$x = \sqrt{20} ÷ \sqrt{5} = \sqrt{20 ÷ 5} = \sqrt{4} = 2$。C
3. 计算：
(1) $\frac{\sqrt{18}}{\sqrt{6}} =$
(2) $\sqrt{\frac{8}{3}} ÷ \sqrt{\frac{1}{2}} =$
(1) $\frac{\sqrt{18}}{\sqrt{6}} =$
$\sqrt{3}$
；(2) $\sqrt{\frac{8}{3}} ÷ \sqrt{\frac{1}{2}} =$
$\frac{4\sqrt{3}}{3}$
。答案：３.(1)$\sqrt{3}$ (2)$\frac{4\sqrt{3}}{3}$
解析：
(1) $\frac{\sqrt{18}}{\sqrt{6}}=\sqrt{\frac{18}{6}}=\sqrt{3}$；
(2) $\sqrt{\frac{8}{3}} ÷ \sqrt{\frac{1}{2}}=\sqrt{\frac{8}{3}÷\frac{1}{2}}=\sqrt{\frac{16}{3}}=\frac{4\sqrt{3}}{3}$。
(2) $\sqrt{\frac{8}{3}} ÷ \sqrt{\frac{1}{2}}=\sqrt{\frac{8}{3}÷\frac{1}{2}}=\sqrt{\frac{16}{3}}=\frac{4\sqrt{3}}{3}$。
4. 化简：
(1) $\sqrt{\frac{5}{49}} =$
(2) $\sqrt{\frac{81}{4}} =$
(3) $\sqrt{\frac{9y}{100x^{4}}} =$
(1) $\sqrt{\frac{5}{49}} =$
$\frac{\sqrt{5}}{7}$
；(2) $\sqrt{\frac{81}{4}} =$
$\frac{9}{2}$
；(3) $\sqrt{\frac{9y}{100x^{4}}} =$
$\frac{3\sqrt{y}}{10x^2}$
。答案：４.(1)$\frac{\sqrt{5}}{7}$ (2)$\frac{9}{2}$ (3)$\frac{3\sqrt{y}}{10x^2}$
5. 计算：
(1) $\sqrt{\frac{1}{24}} ÷ \sqrt{\frac{2}{3}}$；
(2) $\sqrt{12} ÷ \sqrt{27} × \sqrt{18}$；
(3) $\sqrt{\frac{2}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{\frac{5}{7}}$；
(4) $\sqrt{1\frac{3}{5}} × 2\sqrt{3} ÷ (-\frac{1}{2}\sqrt{6})$。
(1) $\sqrt{\frac{1}{24}} ÷ \sqrt{\frac{2}{3}}$；
(2) $\sqrt{12} ÷ \sqrt{27} × \sqrt{18}$；
(3) $\sqrt{\frac{2}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{\frac{5}{7}}$；
(4) $\sqrt{1\frac{3}{5}} × 2\sqrt{3} ÷ (-\frac{1}{2}\sqrt{6})$。
答案：５.(1)$\frac{1}{4}$ (2)$2\sqrt{2}$ (3)$\frac{\sqrt{10}}{5}$ (4)$-\frac{8}{5}\sqrt{5}$
解析：
(1) $\sqrt{\frac{1}{24}} ÷ \sqrt{\frac{2}{3}} = \sqrt{\frac{1}{24} ÷ \frac{2}{3}} = \sqrt{\frac{1}{24} × \frac{3}{2}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$
(2) $\sqrt{12} ÷ \sqrt{27} × \sqrt{18} = \sqrt{\frac{12}{27}} × \sqrt{18} = \sqrt{\frac{4}{9}} × \sqrt{18} = \frac{2}{3} × 3\sqrt{2} = 2\sqrt{2}$
(3) $\sqrt{\frac{2}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{\frac{5}{7}} = \sqrt{\frac{2}{3} ÷ \frac{7}{3} ÷ \frac{5}{7}} = \sqrt{\frac{2}{3} × \frac{3}{7} × \frac{7}{5}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{10}}{5}$
(4) $\sqrt{1\frac{3}{5}} × 2\sqrt{3} ÷ (-\frac{1}{2}\sqrt{6}) = \sqrt{\frac{8}{5}} × 2\sqrt{3} × (-\frac{2}{\sqrt{6}}) = 2\sqrt{\frac{8}{5}} × \sqrt{3} × (-\frac{2}{\sqrt{6}}) = -4\sqrt{\frac{8}{5} × 3 × \frac{1}{6}} = -4\sqrt{\frac{4}{5}} = -4 × \frac{2\sqrt{5}}{5} = -\frac{8}{5}\sqrt{5}$
(2) $\sqrt{12} ÷ \sqrt{27} × \sqrt{18} = \sqrt{\frac{12}{27}} × \sqrt{18} = \sqrt{\frac{4}{9}} × \sqrt{18} = \frac{2}{3} × 3\sqrt{2} = 2\sqrt{2}$
(3) $\sqrt{\frac{2}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{\frac{5}{7}} = \sqrt{\frac{2}{3} ÷ \frac{7}{3} ÷ \frac{5}{7}} = \sqrt{\frac{2}{3} × \frac{3}{7} × \frac{7}{5}} = \sqrt{\frac{2}{5}} = \frac{\sqrt{10}}{5}$
(4) $\sqrt{1\frac{3}{5}} × 2\sqrt{3} ÷ (-\frac{1}{2}\sqrt{6}) = \sqrt{\frac{8}{5}} × 2\sqrt{3} × (-\frac{2}{\sqrt{6}}) = 2\sqrt{\frac{8}{5}} × \sqrt{3} × (-\frac{2}{\sqrt{6}}) = -4\sqrt{\frac{8}{5} × 3 × \frac{1}{6}} = -4\sqrt{\frac{4}{5}} = -4 × \frac{2\sqrt{5}}{5} = -\frac{8}{5}\sqrt{5}$