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1. 二次根式的除法法则：$\frac{\sqrt{a}}{\sqrt{b}} =\_\_\_\_\_\_0$，$b\_\_\_\_\_\_0$)。
答案
1. $\sqrt{\frac{a}{b}}$
2. $\sqrt{\frac{a}{b}} =\_\_\_\_\_\_0$，$b\_\_\_\_\_\_0$)。商的算术平方根，等于的算术平方根除以的算术平方根。
答案
2. $\frac{\sqrt{a}}{\sqrt{b}}$
3. $\sqrt{6} ÷ \sqrt{24} =$
$\frac{1}{2}$
，$\sqrt{\frac{27}{75}} =$$\frac{3}{5}$
，化去分母中二次根式的过程称为“分母有理化
”。答案
3. $\frac{1}{2}$ $\frac{3}{5}$ 分母有理化
4. 计算：$\frac{\sqrt{6}}{\sqrt{15}} =$
$\frac{\sqrt{10}}{5}$
。答案
4. $\frac{\sqrt{10}}{5}$
5. 计算$\frac{\sqrt{5} × \sqrt{12}}{\sqrt{3}}$的结果是
$2\sqrt{5}$
。答案
5. $2\sqrt{5}$
6. 计算：(1)$\sqrt{35} ÷ \sqrt{60}$； (2)$\sqrt{1 \frac{3}{4}} ÷ \sqrt{2 \frac{5}{8}}$； (3)$\sqrt{2 \frac{1}{2}} ÷ 3 \sqrt{28} × (-5 \sqrt{2 \frac{2}{7}})$。
答案
6. 解: (1) $\sqrt{35} ÷ \sqrt{60} = \sqrt{\frac{35}{60}} = \sqrt{\frac{7}{12}} = \frac{\sqrt{21}}{6}$;
(2) $\sqrt{1\frac{3}{4}} ÷ \sqrt{2\frac{5}{8}} = \sqrt{\frac{7}{4} ÷ \frac{21}{8}} = \sqrt{\frac{7}{4} × \frac{8}{21}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$;
(3) $\sqrt{2\frac{1}{2}} ÷ 3\sqrt{28} × (-5\sqrt{2\frac{2}{7}}) = \sqrt{\frac{5}{2}} × \frac{1}{3\sqrt{28}} × (-5\sqrt{\frac{16}{7}}) = -\frac{5}{3}\sqrt{\frac{5}{2} × \frac{1}{28} × \frac{16}{7}} = -\frac{5\sqrt{10}}{21}$.
(2) $\sqrt{1\frac{3}{4}} ÷ \sqrt{2\frac{5}{8}} = \sqrt{\frac{7}{4} ÷ \frac{21}{8}} = \sqrt{\frac{7}{4} × \frac{8}{21}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$;
(3) $\sqrt{2\frac{1}{2}} ÷ 3\sqrt{28} × (-5\sqrt{2\frac{2}{7}}) = \sqrt{\frac{5}{2}} × \frac{1}{3\sqrt{28}} × (-5\sqrt{\frac{16}{7}}) = -\frac{5}{3}\sqrt{\frac{5}{2} × \frac{1}{28} × \frac{16}{7}} = -\frac{5\sqrt{10}}{21}$.
7. 化简$\sqrt{\frac{3a^{2}}{4}}$的结果是(
A.$2 \sqrt{3} a$
B.$\frac{\sqrt{3} |a|}{2}$
C.$\frac{3 \sqrt{a}}{2}$
D.$\frac{\sqrt{3} a}{4}$
B
)A.$2 \sqrt{3} a$
B.$\frac{\sqrt{3} |a|}{2}$
C.$\frac{3 \sqrt{a}}{2}$
D.$\frac{\sqrt{3} a}{4}$
答案
7. B
8. 已知$\sqrt{\frac{1 - a}{a^{2}}} = \frac{\sqrt{1 - a}}{a}$，则$a$的取值范围是(
A.$a ≤ 0$
B.$a ＜ 0$
C.$0 ＜ a ≤ 1$
D.$a ＞ 0$
C
)A.$a ≤ 0$
B.$a ＜ 0$
C.$0 ＜ a ≤ 1$
D.$a ＞ 0$
答案
8. C
9. 化简：(1)$- \sqrt{\frac{1}{27}} =$
$-\frac{\sqrt{3}}{9}$
；(2)$\sqrt{\frac{2}{3}} =$$\frac{\sqrt{6}}{3}$
； (3)$- \sqrt{1 \frac{24}{25}} =$$-\frac{7}{5}$
；(4)$x \sqrt{\frac{1}{8x^{3}}} =$$\frac{\sqrt{2x}}{4x}$
； (5)$\sqrt{\frac{64b^{2}}{9a^{2}}} =$$\frac{8b}{3a}$
($a ＞ 0$，$b ＞ 0$)。答案
9. (1) $-\frac{\sqrt{3}}{9}$ (2) $\frac{\sqrt{6}}{3}$ (3) $-\frac{7}{5}$ (4) $\frac{\sqrt{2x}}{4x}$ (5) $\frac{8b}{3a}$
10. 化简： (1)$\sqrt{\frac{81 × 125}{144}}$；(2)$\sqrt{\frac{121b^{5}}{16a^{2}}}(a ＞ 0)$； (3)$\sqrt{\frac{1}{2b}}$；(4)$\sqrt{\frac{1}{6}}$。
答案
10. 解: (1) $\sqrt{\frac{81 × 125}{144}} = \frac{\sqrt{81 × 125}}{\sqrt{144}} = \frac{9 × 5\sqrt{5}}{12} = \frac{15\sqrt{5}}{4}$.
(2) $\sqrt{\frac{121b^5}{16a^2}} = \frac{\sqrt{121b^4 · b}}{\sqrt{16a^2}} = \frac{11b^2\sqrt{b}}{4a} (a ＞ 0)$.
(3) $\sqrt{\frac{1}{2b}} = \sqrt{\frac{2b}{2b · 2b}} = \sqrt{\frac{2b}{(2b)^2}} = \frac{\sqrt{2b}}{2b}$.
(4) 方法1: $\sqrt{\frac{1}{6}} = \sqrt{\frac{1 × 6}{6 × 6}} = \sqrt{\frac{6}{36}} = \frac{\sqrt{6}}{6}$.
方法2: $\sqrt{\frac{1}{6}} = \frac{\sqrt{1}}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{1 × \sqrt{6}}{\sqrt{6} × \sqrt{6}} = \frac{\sqrt{6}}{6}$.
(2) $\sqrt{\frac{121b^5}{16a^2}} = \frac{\sqrt{121b^4 · b}}{\sqrt{16a^2}} = \frac{11b^2\sqrt{b}}{4a} (a ＞ 0)$.
(3) $\sqrt{\frac{1}{2b}} = \sqrt{\frac{2b}{2b · 2b}} = \sqrt{\frac{2b}{(2b)^2}} = \frac{\sqrt{2b}}{2b}$.
(4) 方法1: $\sqrt{\frac{1}{6}} = \sqrt{\frac{1 × 6}{6 × 6}} = \sqrt{\frac{6}{36}} = \frac{\sqrt{6}}{6}$.
方法2: $\sqrt{\frac{1}{6}} = \frac{\sqrt{1}}{\sqrt{6}} = \frac{1}{\sqrt{6}} = \frac{1 × \sqrt{6}}{\sqrt{6} × \sqrt{6}} = \frac{\sqrt{6}}{6}$.
