1. (2025·如皋期末)下列根式中,是最简二次根式的为 (
A.$\sqrt{2}$
B.$\sqrt{0.5}$
C.$\sqrt{\dfrac{1}{3}}$
D.$\sqrt{12}$
A
)A.$\sqrt{2}$
B.$\sqrt{0.5}$
C.$\sqrt{\dfrac{1}{3}}$
D.$\sqrt{12}$
答案：1. A
2. (教材变式)化简:
(1)$\sqrt{24}=$
(2)$\sqrt{60}=$
(3)$\sqrt{12× 15}=$
(4)$\sqrt{1.8}=$
(5)$\sqrt{\dfrac{2}{3}}=$
(6)$\sqrt{\dfrac{4n^{2}z}{m^{2}}}=$
(1)$\sqrt{24}=$
$2\sqrt{6}$
；(2)$\sqrt{60}=$
$2\sqrt{15}$
；(3)$\sqrt{12× 15}=$
$6\sqrt{5}$
；(4)$\sqrt{1.8}=$
$\frac{3}{5}\sqrt{5}$
；(5)$\sqrt{\dfrac{2}{3}}=$
$\frac{\sqrt{6}}{3}$
；(6)$\sqrt{\dfrac{4n^{2}z}{m^{2}}}=$
$\frac{2n}{m}\sqrt{z}$
.答案：$2. (1) 2\sqrt{6} (2) 2\sqrt{15} (3) 6\sqrt{5} (4) \frac{3}{5}\sqrt{5} (5) \frac{\sqrt{6}}{3} (6) \frac{2n}{m}\sqrt{z}$
3. 使结果中的二次根式为最简二次根式.
(1)$\sqrt{700}$；
(2)$\sqrt{1.6}$；
(3)$\sqrt{\dfrac{9}{2}}$；
(4)$-\sqrt{6\dfrac{2}{3}}$.
(1)$\sqrt{700}$；
(2)$\sqrt{1.6}$；
(3)$\sqrt{\dfrac{9}{2}}$；
(4)$-\sqrt{6\dfrac{2}{3}}$.
答案：$3. (1) 10\sqrt{7} (2) \frac{2}{5}\sqrt{10} (3) \frac{3}{2}\sqrt{2} (4) -\frac{2}{3}\sqrt{15}$
解析：
(1) $\sqrt{700}=\sqrt{100×7}=\sqrt{100}×\sqrt{7}=10\sqrt{7}$；
(2) $\sqrt{1.6}=\sqrt{\dfrac{8}{5}}=\sqrt{\dfrac{4×2×5}{5×5}}=\dfrac{2\sqrt{10}}{5}$；
(3) $\sqrt{\dfrac{9}{2}}=\dfrac{\sqrt{9}}{\sqrt{2}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$；
(4) $-\sqrt{6\dfrac{2}{3}}=-\sqrt{\dfrac{20}{3}}=-\sqrt{\dfrac{20×3}{3×3}}=-\dfrac{2\sqrt{15}}{3}$.
(2) $\sqrt{1.6}=\sqrt{\dfrac{8}{5}}=\sqrt{\dfrac{4×2×5}{5×5}}=\dfrac{2\sqrt{10}}{5}$；
(3) $\sqrt{\dfrac{9}{2}}=\dfrac{\sqrt{9}}{\sqrt{2}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$；
(4) $-\sqrt{6\dfrac{2}{3}}=-\sqrt{\dfrac{20}{3}}=-\sqrt{\dfrac{20×3}{3×3}}=-\dfrac{2\sqrt{15}}{3}$.
4. (易错题)若$xy＞0$,则$\sqrt{-x^{2}y}$可化简为 (
A.$x\sqrt{-y}$
B.$-xy$
C.$-x\sqrt{y}$
D.$-x\sqrt{-y}$
D
)A.$x\sqrt{-y}$
B.$-xy$
C.$-x\sqrt{y}$
D.$-x\sqrt{-y}$
答案：4. D 解析：
∵ xy＞0,
∴ x,y 同号且均不为 0. 又
∵$ \sqrt{-x^{2}y}$有意义,
∴$ -x^{2}y＞0. $
∵$ -x^{2}＜0,$
∴ y＜0.
∴ x＜0.
∴ 原式=|x|$\sqrt{-y}=-x\sqrt{-y}.$
[易错分析]在化简二次根式时,因忽视隐含条件而出错.
∵ xy＞0,
∴ x,y 同号且均不为 0. 又
∵$ \sqrt{-x^{2}y}$有意义,
∴$ -x^{2}y＞0. $
∵$ -x^{2}＜0,$
∴ y＜0.
∴ x＜0.
∴ 原式=|x|$\sqrt{-y}=-x\sqrt{-y}.$
[易错分析]在化简二次根式时,因忽视隐含条件而出错.
5. 计算或化简:
(1)$\sqrt{(\dfrac{1}{4})^{2}+(\dfrac{1}{2})^{2}}=$
(2)$\dfrac{\sqrt{27}}{\sqrt{3x}}=$
(3)$-5\sqrt{3\dfrac{3}{5}}=$
(4)$\dfrac{5ab}{\sqrt{5b}}=$
(1)$\sqrt{(\dfrac{1}{4})^{2}+(\dfrac{1}{2})^{2}}=$
$\frac{\sqrt{5}}{4}$
；(2)$\dfrac{\sqrt{27}}{\sqrt{3x}}=$
$\frac{3}{x}\sqrt{x}$
；(3)$-5\sqrt{3\dfrac{3}{5}}=$
$-3\sqrt{10}$
；(4)$\dfrac{5ab}{\sqrt{5b}}=$
$a\sqrt{5b}$
.答案：$5. (1) \frac{\sqrt{5}}{4} (2) \frac{3}{x}\sqrt{x} (3) -3\sqrt{10} (4) a\sqrt{5b}$
解析：
(1) $\sqrt{(\dfrac{1}{4})^{2}+(\dfrac{1}{2})^{2}}=\sqrt{\dfrac{1}{16}+\dfrac{4}{16}}=\sqrt{\dfrac{5}{16}}=\dfrac{\sqrt{5}}{4}$；
(2) $\dfrac{\sqrt{27}}{\sqrt{3x}}=\sqrt{\dfrac{27}{3x}}=\sqrt{\dfrac{9}{x}}=\dfrac{3}{\sqrt{x}}=\dfrac{3\sqrt{x}}{x}$；
(3) $-5\sqrt{3\dfrac{3}{5}}=-5\sqrt{\dfrac{18}{5}}=-5×\dfrac{3\sqrt{10}}{5}=-3\sqrt{10}$；
(4) $\dfrac{5ab}{\sqrt{5b}}=\dfrac{5ab\sqrt{5b}}{5b}=a\sqrt{5b}$.
(2) $\dfrac{\sqrt{27}}{\sqrt{3x}}=\sqrt{\dfrac{27}{3x}}=\sqrt{\dfrac{9}{x}}=\dfrac{3}{\sqrt{x}}=\dfrac{3\sqrt{x}}{x}$；
(3) $-5\sqrt{3\dfrac{3}{5}}=-5\sqrt{\dfrac{18}{5}}=-5×\dfrac{3\sqrt{10}}{5}=-3\sqrt{10}$；
(4) $\dfrac{5ab}{\sqrt{5b}}=\dfrac{5ab\sqrt{5b}}{5b}=a\sqrt{5b}$.
6. 若长方形的面积为18,一边长为$2\sqrt{2}$,则其邻边长为
$\frac{9}{2}\sqrt{2}$
.答案：$6. \frac{9}{2}\sqrt{2}$
7. (教材变式)计算:
(1)$\sqrt{45}÷ \dfrac{3}{2}\sqrt{\dfrac{1}{5}}× \dfrac{3}{2}\sqrt{5}$；
(2)$-3\sqrt{2\dfrac{1}{3}}× (-\dfrac{1}{8}\sqrt{15})÷ \dfrac{1}{2}\sqrt{\dfrac{2}{5}}$.
(1)$\sqrt{45}÷ \dfrac{3}{2}\sqrt{\dfrac{1}{5}}× \dfrac{3}{2}\sqrt{5}$；
(2)$-3\sqrt{2\dfrac{1}{3}}× (-\dfrac{1}{8}\sqrt{15})÷ \dfrac{1}{2}\sqrt{\dfrac{2}{5}}$.
答案：7. (1) 原式$=(1 × \frac{2}{3} × \frac{3}{2}) × \sqrt{45 × 5 × 5}=15\sqrt{5} (2) $原式$=(3 × \frac{1}{8} × 2) × \sqrt{\frac{7}{3} × 15 × \frac{5}{2}}=\frac{3}{4}\sqrt{\frac{7 × 5 × 5}{2}}=\frac{15}{8}\sqrt{14}$
解析：
(2) 原式$=[(-3)×(-\frac{1}{8})÷\frac{1}{2}]×\sqrt{2\frac{1}{3}×15÷\frac{2}{5}}$
$=(3×\frac{1}{8}×2)×\sqrt{\frac{7}{3}×15×\frac{5}{2}}$
$=\frac{3}{4}×\sqrt{\frac{7×5×5}{2}}$
$=\frac{3}{4}×\frac{5\sqrt{14}}{2}$
$=\frac{15}{8}\sqrt{14}$
$=(3×\frac{1}{8}×2)×\sqrt{\frac{7}{3}×15×\frac{5}{2}}$
$=\frac{3}{4}×\sqrt{\frac{7×5×5}{2}}$
$=\frac{3}{4}×\frac{5\sqrt{14}}{2}$
$=\frac{15}{8}\sqrt{14}$