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1.（2024·大连期末）下列各式是最简二次根式的是 （ ）
A. $\sqrt{0.1}$
B. $\frac{1}{\sqrt{5}}$
C. $\sqrt{27}$
D. $\frac{\sqrt{3}}{3}$
A. $\sqrt{0.1}$
B. $\frac{1}{\sqrt{5}}$
C. $\sqrt{27}$
D. $\frac{\sqrt{3}}{3}$
答案
D
2. 给出下列四道算式：①$\frac{\sqrt{(-4)^{2}ab}}{\sqrt{4ab}}=-4$；②$\frac{\sqrt{3^{2}+4^{2}}}{\sqrt{5^{2}-3^{2}}}=1\frac{1}{4}$；③$\frac{28x}{\sqrt{7x}} = 4\sqrt{x}$；④$\frac{\sqrt{(b - a)^{2}}}{\sqrt{a - b}}=\sqrt{a - b}(a＞b)$.其中正确的算式是 （ ）
A. ①③
B. ②④
C. ①④
D. ②③
A. ①③
B. ②④
C. ①④
D. ②③
答案
B
3. $\frac{2}{\sqrt{5}},\sqrt{\frac{2}{5}},\frac{\sqrt{2}}{5}$的大小关系是 （ ）
A. $\frac{2}{\sqrt{5}}＜\sqrt{\frac{2}{5}}＜\frac{\sqrt{2}}{5}$
B. $\sqrt{\frac{2}{5}}＜\frac{2}{\sqrt{5}}＜\frac{\sqrt{2}}{5}$
C. $\frac{\sqrt{2}}{5}＜\sqrt{\frac{2}{5}}＜\frac{2}{\sqrt{5}}$
D. $\frac{\sqrt{2}}{5}＜\frac{2}{\sqrt{5}}＜\sqrt{\frac{2}{5}}$
A. $\frac{2}{\sqrt{5}}＜\sqrt{\frac{2}{5}}＜\frac{\sqrt{2}}{5}$
B. $\sqrt{\frac{2}{5}}＜\frac{2}{\sqrt{5}}＜\frac{\sqrt{2}}{5}$
C. $\frac{\sqrt{2}}{5}＜\sqrt{\frac{2}{5}}＜\frac{2}{\sqrt{5}}$
D. $\frac{\sqrt{2}}{5}＜\frac{2}{\sqrt{5}}＜\sqrt{\frac{2}{5}}$
答案
C
4. 化简：(1)$\sqrt{\frac{16}{3}}=$_______；
(2)$\sqrt{1.5}=$_______；(3)$\frac{\sqrt{75}}{\sqrt{8}}=$_______.
(2)$\sqrt{1.5}=$_______；(3)$\frac{\sqrt{75}}{\sqrt{8}}=$_______.
答案
(1)$\frac{4\sqrt{3}}{3}$ (2)$\frac{\sqrt{6}}{2}$ (3)$\frac{5\sqrt{6}}{4}$
5. 设长方形的面积是$S$，相邻两边的长分别是$a、b$.
(1)若$a = \sqrt{8},b = \sqrt{12}$，则$S =$_______；
(2)若$S = 16,a = \sqrt{6}$，则$b =$_______.
(1)若$a = \sqrt{8},b = \sqrt{12}$，则$S =$_______；
(2)若$S = 16,a = \sqrt{6}$，则$b =$_______.
答案
(1)$4\sqrt{6}$ (2)$\frac{8\sqrt{6}}{3}$
6. 把下列各式化成最简二次根式：
(1)$\sqrt{\frac{1}{12}}$； (2)$\sqrt{1\frac{4}{5}}$；
(3)$\frac{2x}{\sqrt{3x}}(x＞0)$； (4)$\sqrt{\frac{3b}{8a^{3}}}(a＞0,b＞0)$.
(1)$\sqrt{\frac{1}{12}}$； (2)$\sqrt{1\frac{4}{5}}$；
(3)$\frac{2x}{\sqrt{3x}}(x＞0)$； (4)$\sqrt{\frac{3b}{8a^{3}}}(a＞0,b＞0)$.
答案
(1)$\frac{\sqrt{3}}{6}$ (2)$\frac{3\sqrt{5}}{5}$ (3)$\frac{2\sqrt{3x}}{3}$ (4)$\frac{\sqrt{6ab}}{4a^{2}}$
7. 计算：
(1)$\sqrt{1\frac{1}{3}}\div\sqrt{2\frac{1}{3}}\div\sqrt{1\frac{2}{5}}$；
(2)$\sqrt{\frac{b}{a}}\div\sqrt{ab}\cdot\sqrt{\frac{a^{3}}{b}}(a＞0,b＞0)$；
(3)$\sqrt{18}\div(\sqrt{84}\times\sqrt{\frac{2}{7}})$；
(4)$8x^{2}\sqrt{xy}\div\frac{1}{3}\sqrt{\frac{x^{3}}{y}}\times(-\sqrt{\frac{y^{2}}{x}})(x＞0,y＞0)$.
(1)$\sqrt{1\frac{1}{3}}\div\sqrt{2\frac{1}{3}}\div\sqrt{1\frac{2}{5}}$；
(2)$\sqrt{\frac{b}{a}}\div\sqrt{ab}\cdot\sqrt{\frac{a^{3}}{b}}(a＞0,b＞0)$；
(3)$\sqrt{18}\div(\sqrt{84}\times\sqrt{\frac{2}{7}})$；
(4)$8x^{2}\sqrt{xy}\div\frac{1}{3}\sqrt{\frac{x^{3}}{y}}\times(-\sqrt{\frac{y^{2}}{x}})(x＞0,y＞0)$.
答案
(1)原式 = $\sqrt{\frac{4}{3} \times \frac{3}{7} \times \frac{5}{7}} = \sqrt{\frac{20}{7^{2}}} = \frac{2\sqrt{5}}{7}$.
(2)原式 = $\sqrt{\frac{b}{a} \cdot \frac{1}{ab} \cdot \frac{a^{3}}{b}} = \sqrt{\frac{a}{b}} = \frac{\sqrt{ab}}{b}$.
(3)原式 = $\sqrt{18} \div \sqrt{84 \times \frac{2}{7}} = \sqrt{18} \div \sqrt{24} = \sqrt{\frac{18}{24}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
(4)原式 = $8x^{2}\sqrt{xy} \div \sqrt{\frac{x^{3}}{9y}} \times (-\sqrt{\frac{y^{2}}{x}}) = -8x^{2}\sqrt{xy \cdot \frac{9y}{x^{3}} \cdot \frac{y^{2}}{x}} = -8x^{2}\sqrt{\frac{9y^{4}}{x^{3}}} = -8x^{2} \cdot \frac{3y^{2}\sqrt{x}}{x^{2}} = -24y^{2}\sqrt{x}$.
(2)原式 = $\sqrt{\frac{b}{a} \cdot \frac{1}{ab} \cdot \frac{a^{3}}{b}} = \sqrt{\frac{a}{b}} = \frac{\sqrt{ab}}{b}$.
(3)原式 = $\sqrt{18} \div \sqrt{84 \times \frac{2}{7}} = \sqrt{18} \div \sqrt{24} = \sqrt{\frac{18}{24}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
(4)原式 = $8x^{2}\sqrt{xy} \div \sqrt{\frac{x^{3}}{9y}} \times (-\sqrt{\frac{y^{2}}{x}}) = -8x^{2}\sqrt{xy \cdot \frac{9y}{x^{3}} \cdot \frac{y^{2}}{x}} = -8x^{2}\sqrt{\frac{9y^{4}}{x^{3}}} = -8x^{2} \cdot \frac{3y^{2}\sqrt{x}}{x^{2}} = -24y^{2}\sqrt{x}$.
8. 计算$\sqrt{\frac{a}{b}}\div\sqrt{ab}\cdot\sqrt{\frac{1}{ab}}(a＜0,b＜0)$的结果是 （ ）
A. $\frac{1}{b}\sqrt{ab}$
B. $-\frac{1}{b}\sqrt{ab}$
C. $\frac{1}{ab^{2}}\sqrt{ab}$
D. $-\frac{1}{ab^{2}}\sqrt{ab}$
A. $\frac{1}{b}\sqrt{ab}$
B. $-\frac{1}{b}\sqrt{ab}$
C. $\frac{1}{ab^{2}}\sqrt{ab}$
D. $-\frac{1}{ab^{2}}\sqrt{ab}$
答案
D 解析：$\sqrt{\frac{a}{b}} \div \sqrt{ab} \cdot \sqrt{\frac{1}{ab}} = \sqrt{\frac{a}{b} \times \frac{1}{ab} \times \frac{1}{ab}} = \sqrt{\frac{1}{ab^{3}}} = \frac{\sqrt{ab}}{\sqrt{a^{2}b^{4}}} = -\frac{1}{ab^{2}}\sqrt{ab}$. 故选 D.
