1. (2025·无锡校级月考)下列二次根式中,属于最简二次根式的是 (
A
)
A.$\sqrt{x^{2}+y^{2}}$
B.$\sqrt{8}$
C.$\sqrt{\dfrac{1}{3}}$
D.$\sqrt{0.5}$
答案：1. A 解析：A.无法继续化简，属于最简二次根式；B.$\sqrt{8}=2\sqrt{2}$，原式不属于最简二次根式；C.$\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt{3}}{3}$，原式不属于最简二次根式；D.$\sqrt{0.5}=\sqrt{\dfrac{1}{2}}=\dfrac{\sqrt{2}}{2}$，原式不属于最简二次根式.故选A.
2. 给出下列四道算式:①$\dfrac{\sqrt{(-
4)^{2}ab}}{\sqrt{4ab}}=-4$;②$\dfrac{\sqrt{
3^{2}+4^{2}}}{\sqrt{5^{2}-3^{2}}}=1\dfrac{1}{4}$;③$\dfrac{28x}{\sqrt{7x}}=4\sqrt{x}$;④$\dfrac{\sqrt{(b-a)^{2}}}{\sqrt{a-b}}=\sqrt{a-b}(a＞b)$.其中正确的算式是 (
B
)
A.①③
B.②④
C.①④
D.②③
答案：2. B 解析：①$\dfrac{\sqrt{(-4)^{2}ab}}{\sqrt{4ab}}=\sqrt{\dfrac{16ab}{4ab}}=2$，故①错误；②$\dfrac{\sqrt{3^{2}+4^{2}}}{\sqrt{5^{2}-3^{2}}}=\sqrt{\dfrac{5^{2}}{4^{2}}}=1\dfrac{1}{4}$，故②正确；③$\dfrac{28x}{\sqrt{7x}}=\dfrac{28x·\sqrt{7x}}{\sqrt{7x}·\sqrt{7x}}=\dfrac{28x·\sqrt{7x}}{7x}=4\sqrt{7x}$，故③错误；④$\dfrac{\sqrt{(b - a)^{2}}}{\sqrt{a - b}}=\dfrac{a - b}{\sqrt{a - b}}=\sqrt{a - b}$，故④正确.故选B.
3. 一题多解 $\dfrac{2}{\sqrt{5}}$,$\sqrt{\dfrac{2}{5}}$,$\dfrac{\sqrt{2}}{5}$的大小关系是 (
C
)
A.$\dfrac{2}{\sqrt{5}}＜\sqrt{\dfrac{2}{5}}＜\dfrac{\sqrt{2}}{5}$
B.$\sqrt{\dfrac{2}{5}}＜\dfrac{2}{\sqrt{5}}＜\dfrac{\sqrt{2}}{5}$
C.$\dfrac{\sqrt{2}}{5}＜\sqrt{\dfrac{2}{5}}＜\dfrac{2}{\sqrt{5}}$
D.$\dfrac{\sqrt{2}}{5}＜\dfrac{2}{\sqrt{5}}＜\sqrt{\dfrac{2}{5}}$
答案：3. C 解析：$\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}$，$\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}$，$\because\sqrt{2}＜\sqrt{10}＜2\sqrt{5}$，$\therefore\dfrac{\sqrt{2}}{5}＜\sqrt{\dfrac{2}{5}}＜\dfrac{2}{\sqrt{5}}$.
一题多解
$\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{2}}{\sqrt{5}}$，对于$\dfrac{2}{\sqrt{5}}$和$\dfrac{\sqrt{2}}{\sqrt{5}}$，分母相同而分子不同，比较可得$\dfrac{\sqrt{2}}{\sqrt{5}}＜\dfrac{2}{\sqrt{5}}$；对于$\dfrac{\sqrt{2}}{\sqrt{5}}$和$\dfrac{\sqrt{2}}{5}$，分子相同而分母不同，比较可得$\dfrac{\sqrt{2}}{5}＜\dfrac{\sqrt{2}}{\sqrt{5}}$，$\therefore\dfrac{\sqrt{2}}{5}＜\sqrt{\dfrac{2}{5}}＜\dfrac{2}{\sqrt{5}}$.
4. 化简:(1)$\sqrt{\dfrac{16}{3}}=$
$\dfrac{4\sqrt{3}}{3}$
;
(2)$\sqrt{1.5}=$
$\dfrac{\sqrt{6}}{2}$
;(3)$\dfrac{\sqrt{75}}{\sqrt{8}}=$
$\dfrac{5\sqrt{6}}{4}$
.
答案：4. （1）$\dfrac{4\sqrt{3}}{3}$ 解析：原式$=\dfrac{\sqrt{16×3}}{3}=\dfrac{4\sqrt{3}}{3}$.
（2）$\dfrac{\sqrt{6}}{2}$ 解析：原式$=\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{3×2}}{2}=\dfrac{\sqrt{6}}{2}$.
（3）$\dfrac{5\sqrt{6}}{4}$ 解析：原式$=\dfrac{5\sqrt{3}}{2\sqrt{2}}=\dfrac{5\sqrt{3×2}}{2×2}=\dfrac{5\sqrt{6}}{4}$.
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$S=$
$4\sqrt{6}$
;
(2)若$S=16$,$a=\sqrt{6}$,则$b=$
$\dfrac{8\sqrt{6}}{3}$
.
答案：5. （1）$4\sqrt{6}$ 解析：$S = ab=\sqrt{8}×\sqrt{12}=2\sqrt{2}×2\sqrt{3}=4\sqrt{6}$.
（2）$\dfrac{8\sqrt{6}}{3}$ 解析：$b=\dfrac{S}{a}=\dfrac{16}{\sqrt{6}}=\dfrac{16\sqrt{6}}{6}=\dfrac{8\sqrt{6}}{3}$.
6. 把下列各式化成最简二次根式:
(1)$\sqrt{\dfrac{1}{12}}$;
(2)$\sqrt{1\dfrac{4}{5}}$;
(3)$\dfrac{2x}{\sqrt{3x}}(x＞0)$;
(4)$\sqrt{\dfrac{3b}{8a^{3}}}(a＞0,b＞0)$.
答案：6. （1）原式$=\dfrac{\sqrt{12}}{12}=\dfrac{2\sqrt{3}}{12}=\dfrac{\sqrt{3}}{6}$.
（2）原式$=\sqrt{\dfrac{9}{5}}=\dfrac{3\sqrt{5}}{5}$.
（3）原式$=\dfrac{2x\sqrt{3x}}{3x}=\dfrac{2\sqrt{3x}}{3}$.
（4）原式$=\dfrac{\sqrt{3b·8a^{3}}}{8a^{3}}=\dfrac{2a\sqrt{6ab}}{8a^{3}}=\dfrac{\sqrt{6ab}}{4a^{2}}$.
7. 计算:
(1)$\sqrt{1\dfrac{1}{3}}÷\sqrt{2\dfrac{1}{3}}÷\sqrt{1\dfrac{2}{5}}$;
(2)$\sqrt{\dfrac{b}{a}}÷\sqrt{ab}·\sqrt{\dfrac{a^{3}}{b}}(a＞0,b＞0)$;
(3)$\sqrt{18}÷(\sqrt{84}×\sqrt{\dfrac{2}{7}})$;
(4)$8x^{2}\sqrt{xy}÷\dfrac{1}{3}\sqrt{\dfrac{x^{3}}{y}}·(-\sqrt{\dfrac{y^{2}}{x}})(x＞0,y＞0)$.
答案：7. （1）原式$=\sqrt{\dfrac{4}{3}×\dfrac{3}{7}×\dfrac{5}{7}}=\sqrt{\dfrac{20}{7^{2}}}=\dfrac{2\sqrt{5}}{7}$.
（2）原式$=\sqrt{\dfrac{b}{a}·\dfrac{1}{ab}·\dfrac{a^{3}}{b}}=\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{ab}}{b}$.
（3）原式$=\sqrt{18}÷\sqrt{84×\dfrac{2}{7}}=\sqrt{18}÷\sqrt{24}=\sqrt{\dfrac{18}{24}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2}$.
（4）原式$=8x^{2}\sqrt{xy}÷\sqrt{\dfrac{x^{3}}{9y}}·(-\sqrt{\dfrac{y^{2}}{x}})=-8x^{2}·\sqrt{xy·\dfrac{9y}{x^{3}}·\dfrac{y^{2}}{x}}=-8x^{2}\sqrt{\dfrac{9y^{4}}{x^{3}}}=-8x^{2}·\dfrac{3y^{2}\sqrt{x}}{x^{2}}=-24y^{2}\sqrt{x}$.
8. 计算$\sqrt{\dfrac{a}{b}}÷\sqrt{ab}·\sqrt{\dfrac{1}{ab}}(a＜0,b＜0)$的结果是 (
D
)
A.$\dfrac{1}{b}\sqrt{ab}$
B.$-\dfrac{1}{b}\sqrt{ab}$
C.$\dfrac{1}{ab^{2}}\sqrt{ab}$
D.$-\dfrac{1}{ab^{2}}\sqrt{ab}$
答案：8. D 解析：$\sqrt{\dfrac{a}{b}}÷\sqrt{ab}·\sqrt{\dfrac{1}{ab}}=\sqrt{\dfrac{a}{b}×\dfrac{1}{ab}×\dfrac{1}{ab}}=\sqrt{\dfrac{1}{ab^{3}}}=\sqrt{\dfrac{ab}{a^{2}b^{4}}}=-\dfrac{1}{ab^{2}}\sqrt{ab}$.故选D.
