1.(2024·秦淮区期末)下列二次根式中,最简二次根式是( )
A. $\sqrt{12}$
B. $\sqrt{14}$
C. $\sqrt{1.6}$
D. $\sqrt{\frac{1}{2}}$
A. $\sqrt{12}$
B. $\sqrt{14}$
C. $\sqrt{1.6}$
D. $\sqrt{\frac{1}{2}}$
答案
B
2. 将$\sqrt{\frac{45}{2}}$化为最简二次根式,其结果是( )
A. $\frac{\sqrt{45}}{2}$
B. $\frac{\sqrt{90}}{2}$
C. $\frac{9\sqrt{10}}{2}$
D. $\frac{3\sqrt{10}}{2}$
A. $\frac{\sqrt{45}}{2}$
B. $\frac{\sqrt{90}}{2}$
C. $\frac{9\sqrt{10}}{2}$
D. $\frac{3\sqrt{10}}{2}$
答案
D
3. 下列各式中,化简正确的是( )
A. $\sqrt{\frac{5}{3}} = 3\sqrt{15}$
B. $\sqrt{\frac{1}{2}} = \pm\frac{1}{2}\sqrt{2}$
C. $\sqrt{a^{4}b} = a^{2}\sqrt{b}$
D. $\sqrt{x^{3}-x^{2}} = -x\sqrt{x - 1}$
A. $\sqrt{\frac{5}{3}} = 3\sqrt{15}$
B. $\sqrt{\frac{1}{2}} = \pm\frac{1}{2}\sqrt{2}$
C. $\sqrt{a^{4}b} = a^{2}\sqrt{b}$
D. $\sqrt{x^{3}-x^{2}} = -x\sqrt{x - 1}$
答案
C
4.(2023·亭湖区期末)计算$\sqrt{2} \times \sqrt{3} \div \frac{1}{\sqrt{6}}$的结果为________.
答案
6
5.(2023·海陵区期末)若$\sqrt{3a + 1}$是最简二次根式,且$a$为整数,则$a$的最小值是________.
答案
2
6. 把下列各式化成最简二次根式.
(1)$\sqrt{\frac{49a^{3}}{9}}$; (2)$\sqrt{\frac{1}{4} + \frac{1}{9}}$; (3)$x^{2}\sqrt{\frac{2}{x}}$.
(1)$\sqrt{\frac{49a^{3}}{9}}$; (2)$\sqrt{\frac{1}{4} + \frac{1}{9}}$; (3)$x^{2}\sqrt{\frac{2}{x}}$.
答案
(1) $\frac{7}{3}a\sqrt{a}$
(2) $\frac{\sqrt{13}}{6}$
(3) $x\sqrt{2x}$
(2) $\frac{\sqrt{13}}{6}$
(3) $x\sqrt{2x}$
7. 计算:
(1)$\sqrt{12} \div \sqrt{\frac{27}{2}}$; (2)$\sqrt{6x^{2}} \div \sqrt{12x^{3}y}(y > 0)$;
(3)$-6\sqrt{8} \times 2\sqrt{6} \div 4\sqrt{27}$; (4)$\sqrt{30} \div 3\sqrt{\frac{8}{5}} \times \frac{2}{3}\sqrt{\frac{20}{3}}$.
(1)$\sqrt{12} \div \sqrt{\frac{27}{2}}$; (2)$\sqrt{6x^{2}} \div \sqrt{12x^{3}y}(y > 0)$;
(3)$-6\sqrt{8} \times 2\sqrt{6} \div 4\sqrt{27}$; (4)$\sqrt{30} \div 3\sqrt{\frac{8}{5}} \times \frac{2}{3}\sqrt{\frac{20}{3}}$.
答案
解:(1) 原式$=\sqrt{12\div\frac{27}{2}}=\sqrt{12\times\frac{2}{27}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$.
(2) 原式$=\sqrt{\frac{6x^{2}}{12x^{3}y}}=\sqrt{\frac{1}{2xy}}=\frac{\sqrt{2xy}}{2xy}$.
(3) 原式$=-12\sqrt{2}\times2\sqrt{6}\div12\sqrt{3}=-48\sqrt{3}\div12\sqrt{3}=-4$.
(4) 原式$=(1\times\frac{1}{3}\times\frac{2}{3})\sqrt{30\times\frac{5}{8}\times\frac{20}{3}}=\frac{2}{9}\times5\sqrt{5}=\frac{10}{9}\sqrt{5}$.
(2) 原式$=\sqrt{\frac{6x^{2}}{12x^{3}y}}=\sqrt{\frac{1}{2xy}}=\frac{\sqrt{2xy}}{2xy}$.
(3) 原式$=-12\sqrt{2}\times2\sqrt{6}\div12\sqrt{3}=-48\sqrt{3}\div12\sqrt{3}=-4$.
(4) 原式$=(1\times\frac{1}{3}\times\frac{2}{3})\sqrt{30\times\frac{5}{8}\times\frac{20}{3}}=\frac{2}{9}\times5\sqrt{5}=\frac{10}{9}\sqrt{5}$.
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