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6. 下列根式中是最简二次根式的是(
A.$\sqrt{\frac{2}{3}}$
B.$\sqrt{3}$
C.$\sqrt{9}$
D.$\sqrt{12}$
B
)A.$\sqrt{\frac{2}{3}}$
B.$\sqrt{3}$
C.$\sqrt{9}$
D.$\sqrt{12}$
答案
6. B
7. 把$m\sqrt{-\frac{1}{m}}$根号外的因式移入根号内得(
A.$\sqrt{m}$
B.$\sqrt{-m}$
C.$-\sqrt{m}$
D.$-\sqrt{-m}$
D
)A.$\sqrt{m}$
B.$\sqrt{-m}$
C.$-\sqrt{m}$
D.$-\sqrt{-m}$
答案
7. D
解析
要使$m\sqrt{-\frac{1}{m}}$有意义，则$-\frac{1}{m} ＞ 0$，即$m ＜ 0$。
$m\sqrt{-\frac{1}{m}} = -|m|\sqrt{-\frac{1}{m}} = -\sqrt{|m|^2 · (-\frac{1}{m})} = -\sqrt{m^2 · (-\frac{1}{m})} = -\sqrt{-m}$
D
$m\sqrt{-\frac{1}{m}} = -|m|\sqrt{-\frac{1}{m}} = -\sqrt{|m|^2 · (-\frac{1}{m})} = -\sqrt{m^2 · (-\frac{1}{m})} = -\sqrt{-m}$
D
8. 计算：
（1）$\sqrt{27} × \sqrt{50} ÷ \sqrt{6}$；
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$；
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$.
（1）$\sqrt{27} × \sqrt{50} ÷ \sqrt{6}$；
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$；
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$.
答案
（1）$\sqrt{27} × \sqrt{50} ÷ \sqrt{6}$
$=3\sqrt{3}×5\sqrt{2}÷\sqrt{6}$
$=15\sqrt{6}÷\sqrt{6}$
$=15$
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$
$=\sqrt{\frac{4}{3}}÷\sqrt{\frac{8}{3}}×\sqrt{\frac{8}{5}}$
$=\sqrt{\frac{4}{3}÷\frac{8}{3}×\frac{8}{5}}$
$=\sqrt{\frac{4}{3}×\frac{3}{8}×\frac{8}{5}}$
$=\sqrt{\frac{4}{5}}$
$=\frac{2\sqrt{5}}{5}$
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$
$=2×2\sqrt{3}÷(\frac{1}{2}×5\sqrt{2})×(\frac{1}{2}×\frac{\sqrt{3}}{2})$
$=4\sqrt{3}÷\frac{5\sqrt{2}}{2}×\frac{\sqrt{3}}{4}$
$=4\sqrt{3}×\frac{2}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{8\sqrt{3}}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{2\sqrt{3}×\sqrt{3}}{5\sqrt{2}}$
$=\frac{6}{5\sqrt{2}}$
$=\frac{3\sqrt{2}}{5}$
$=3\sqrt{3}×5\sqrt{2}÷\sqrt{6}$
$=15\sqrt{6}÷\sqrt{6}$
$=15$
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$
$=\sqrt{\frac{4}{3}}÷\sqrt{\frac{8}{3}}×\sqrt{\frac{8}{5}}$
$=\sqrt{\frac{4}{3}÷\frac{8}{3}×\frac{8}{5}}$
$=\sqrt{\frac{4}{3}×\frac{3}{8}×\frac{8}{5}}$
$=\sqrt{\frac{4}{5}}$
$=\frac{2\sqrt{5}}{5}$
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$
$=2×2\sqrt{3}÷(\frac{1}{2}×5\sqrt{2})×(\frac{1}{2}×\frac{\sqrt{3}}{2})$
$=4\sqrt{3}÷\frac{5\sqrt{2}}{2}×\frac{\sqrt{3}}{4}$
$=4\sqrt{3}×\frac{2}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{8\sqrt{3}}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{2\sqrt{3}×\sqrt{3}}{5\sqrt{2}}$
$=\frac{6}{5\sqrt{2}}$
$=\frac{3\sqrt{2}}{5}$
解析
（1）$\sqrt{27} × \sqrt{50} ÷ \sqrt{6}$
$=3\sqrt{3}×5\sqrt{2}÷\sqrt{6}$
$=15\sqrt{6}÷\sqrt{6}$
$=15$
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$
$=\sqrt{\frac{4}{3}}÷\sqrt{\frac{8}{3}}×\sqrt{\frac{8}{5}}$
$=\sqrt{\frac{4}{3}÷\frac{8}{3}×\frac{8}{5}}$
$=\sqrt{\frac{4}{3}×\frac{3}{8}×\frac{8}{5}}$
$=\sqrt{\frac{4}{5}}$
$=\frac{2\sqrt{5}}{5}$
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$
$=2×2\sqrt{3}÷(\frac{1}{2}×5\sqrt{2})×(\frac{1}{2}×\frac{\sqrt{3}}{2})$
$=4\sqrt{3}÷\frac{5\sqrt{2}}{2}×\frac{\sqrt{3}}{4}$
$=4\sqrt{3}×\frac{2}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{8\sqrt{3}}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{2\sqrt{3}×\sqrt{3}}{5\sqrt{2}}$
$=\frac{6}{5\sqrt{2}}$
$=\frac{3\sqrt{2}}{5}$
$=3\sqrt{3}×5\sqrt{2}÷\sqrt{6}$
$=15\sqrt{6}÷\sqrt{6}$
$=15$
（2）$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{2}{3}} × \sqrt{1\frac{3}{5}}$
$=\sqrt{\frac{4}{3}}÷\sqrt{\frac{8}{3}}×\sqrt{\frac{8}{5}}$
$=\sqrt{\frac{4}{3}÷\frac{8}{3}×\frac{8}{5}}$
$=\sqrt{\frac{4}{3}×\frac{3}{8}×\frac{8}{5}}$
$=\sqrt{\frac{4}{5}}$
$=\frac{2\sqrt{5}}{5}$
（3）$2\sqrt{12} ÷ \frac{1}{2}\sqrt{50} × \frac{1}{2}\sqrt{\frac{3}{4}}$
$=2×2\sqrt{3}÷(\frac{1}{2}×5\sqrt{2})×(\frac{1}{2}×\frac{\sqrt{3}}{2})$
$=4\sqrt{3}÷\frac{5\sqrt{2}}{2}×\frac{\sqrt{3}}{4}$
$=4\sqrt{3}×\frac{2}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{8\sqrt{3}}{5\sqrt{2}}×\frac{\sqrt{3}}{4}$
$=\frac{2\sqrt{3}×\sqrt{3}}{5\sqrt{2}}$
$=\frac{6}{5\sqrt{2}}$
$=\frac{3\sqrt{2}}{5}$
9. 化简：
（1）$\sqrt{xy} ÷ 2\sqrt{\frac{y^2}{x}}$；
（2）$\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}$；
（3）$\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}$.
（1）$\sqrt{xy} ÷ 2\sqrt{\frac{y^2}{x}}$；
（2）$\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}$；
（3）$\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}$.
答案
9.(1) $\sqrt{xy} ÷ 2\sqrt{\frac{y^2}{x}}=\frac{1}{2}\sqrt{xy÷\frac{y^2}{x}}=\frac{1}{2}\sqrt{xy×\frac{x}{y^2}}=\frac{1}{2}\sqrt{\frac{x^2}{y}}=\frac{1}{2}×\frac{x}{\sqrt{y}}=\frac{x\sqrt{y}}{2y}$
(2) $\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}=\sqrt{18÷8÷\frac{27}{2}}=\sqrt{18×\frac{1}{8}×\frac{2}{27}}=\sqrt{\frac{36}{216}}=\sqrt{\frac{1}{6}}=\frac{\sqrt{6}}{6}$
(3) $\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}=\sqrt{2}×\frac{\sqrt{6}}{3}×\frac{\sqrt{3}}{4}=\frac{\sqrt{2}×\sqrt{6}×\sqrt{3}}{12}=\frac{\sqrt{36}}{12}=\frac{6}{12}=\frac{1}{2}$
(2) $\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}=\sqrt{18÷8÷\frac{27}{2}}=\sqrt{18×\frac{1}{8}×\frac{2}{27}}=\sqrt{\frac{36}{216}}=\sqrt{\frac{1}{6}}=\frac{\sqrt{6}}{6}$
(3) $\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}=\sqrt{2}×\frac{\sqrt{6}}{3}×\frac{\sqrt{3}}{4}=\frac{\sqrt{2}×\sqrt{6}×\sqrt{3}}{12}=\frac{\sqrt{36}}{12}=\frac{6}{12}=\frac{1}{2}$
解析
(1) $\sqrt{xy} ÷ 2\sqrt{\frac{y^2}{x}}=\frac{1}{2}\sqrt{xy÷\frac{y^2}{x}}=\frac{1}{2}\sqrt{xy×\frac{x}{y^2}}=\frac{1}{2}\sqrt{\frac{x^2}{y}}=\frac{1}{2}×\frac{x}{\sqrt{y}}=\frac{x\sqrt{y}}{2y}$
(2) $\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}=\sqrt{18÷8÷\frac{27}{2}}=\sqrt{18×\frac{1}{8}×\frac{2}{27}}=\sqrt{\frac{36}{216}}=\sqrt{\frac{1}{6}}=\frac{\sqrt{6}}{6}$
(3) $\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}=\sqrt{2}×\frac{\sqrt{6}}{3}×\frac{\sqrt{3}}{4}=\frac{\sqrt{2}×\sqrt{6}×\sqrt{3}}{12}=\frac{\sqrt{36}}{12}=\frac{6}{12}=\frac{1}{2}$
(2) $\sqrt{18} ÷ \sqrt{8} ÷ \sqrt{\frac{27}{2}}=\sqrt{18÷8÷\frac{27}{2}}=\sqrt{18×\frac{1}{8}×\frac{2}{27}}=\sqrt{\frac{36}{216}}=\sqrt{\frac{1}{6}}=\frac{\sqrt{6}}{6}$
(3) $\sqrt{2} × \frac{\sqrt{6}}{3} ÷ \frac{4}{\sqrt{3}}=\sqrt{2}×\frac{\sqrt{6}}{3}×\frac{\sqrt{3}}{4}=\frac{\sqrt{2}×\sqrt{6}×\sqrt{3}}{12}=\frac{\sqrt{36}}{12}=\frac{6}{12}=\frac{1}{2}$
10. $\frac{3}{5}\sqrt{xy^2} ÷ (-\frac{4}{15}\sqrt{\frac{y}{x}}) · (-\frac{5}{6}\sqrt{x^3y})$.
答案
$\begin{aligned}&\frac{3}{5}\sqrt{xy^2} ÷ (-\frac{4}{15}\sqrt{\frac{y}{x}}) · (-\frac{5}{6}\sqrt{x^3y})\\=&\frac{3}{5}y\sqrt{x} ÷ (-\frac{4}{15}·\frac{\sqrt{xy}}{x}) · (-\frac{5}{6}x\sqrt{xy})\\=&\frac{3}{5}y\sqrt{x} × (-\frac{15x}{4\sqrt{xy}}) × (-\frac{5}{6}x\sqrt{xy})\\=&\frac{3}{5}y\sqrt{x} × \frac{15x}{4\sqrt{xy}} × \frac{5}{6}x\sqrt{xy}\\=&\frac{3}{5} × \frac{15x}{4} × \frac{5}{6}x × y\sqrt{x} × \frac{1}{\sqrt{xy}} × \sqrt{xy}\\=&\frac{3 × 15x × 5x}{5 × 4 × 6} × y\sqrt{x}\\=&\frac{225x^2}{120}y\sqrt{x}\\=&\frac{15}{8}x^2y\sqrt{x}\end{aligned}$
解析
$\begin{aligned}&\frac{3}{5}\sqrt{xy^2} ÷ (-\frac{4}{15}\sqrt{\frac{y}{x}}) · (-\frac{5}{6}\sqrt{x^3y})\\=&\frac{3}{5}y\sqrt{x} ÷ (-\frac{4}{15}·\frac{\sqrt{xy}}{x}) · (-\frac{5}{6}x\sqrt{xy})\\=&\frac{3}{5}y\sqrt{x} × (-\frac{15x}{4\sqrt{xy}}) × (-\frac{5}{6}x\sqrt{xy})\\=&\frac{3}{5}y\sqrt{x} × \frac{15x}{4\sqrt{xy}} × \frac{5}{6}x\sqrt{xy}\\=&\frac{3}{5} × \frac{15x}{4} × \frac{5}{6}x × y\sqrt{x} × \frac{1}{\sqrt{xy}} × \sqrt{xy}\\=&\frac{3 × 15x × 5x}{5 × 4 × 6} × y\sqrt{x}\\=&\frac{225x^2}{120}y\sqrt{x}\\=&\frac{15}{8}x^2y\sqrt{x}\end{aligned}$
