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1. 化简$\sqrt{8} × \sqrt{2}$的结果为(
A.$\sqrt{16}$
B.$\sqrt{4}$
C.4
D.16
C
)A.$\sqrt{16}$
B.$\sqrt{4}$
C.4
D.16
答案
C
解析
根据二次根式的乘法法则，有$\sqrt{a} × \sqrt{b} = \sqrt{a × b}$（其中$a \geq 0, b \geq 0$）。
应用这一法则，计算$\sqrt{8} × \sqrt{2}$：
$\sqrt{8} × \sqrt{2} = \sqrt{8 × 2} = \sqrt{16} = 4$
应用这一法则，计算$\sqrt{8} × \sqrt{2}$：
$\sqrt{8} × \sqrt{2} = \sqrt{8 × 2} = \sqrt{16} = 4$
2. 下列各等式成立的是(
A.$4\sqrt{5} × 2\sqrt{5} = 8\sqrt{5}$
B.$5\sqrt{5} × 4\sqrt{5} = 20\sqrt{10}$
C.$4\sqrt{3} × 3\sqrt{2} = 7\sqrt{5}$
D.$5\sqrt{3} × 4\sqrt{2} = 20\sqrt{6}$
D
)A.$4\sqrt{5} × 2\sqrt{5} = 8\sqrt{5}$
B.$5\sqrt{5} × 4\sqrt{5} = 20\sqrt{10}$
C.$4\sqrt{3} × 3\sqrt{2} = 7\sqrt{5}$
D.$5\sqrt{3} × 4\sqrt{2} = 20\sqrt{6}$
答案
D
解析
A. $4\sqrt{5}×2\sqrt{5}=4×2×(\sqrt{5}×\sqrt{5})=8×5=40≠8\sqrt{5}$；B. $5\sqrt{5}×4\sqrt{5}=5×4×(\sqrt{5}×\sqrt{5})=20×5=100≠20\sqrt{10}$；C. $4\sqrt{3}×3\sqrt{2}=4×3×\sqrt{3×2}=12\sqrt{6}≠7\sqrt{5}$；D. $5\sqrt{3}×4\sqrt{2}=5×4×\sqrt{3×2}=20\sqrt{6}$，成立。
3. 计算：$\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{1\frac{2}{5}} = $(
A.$\frac{2\sqrt{5}}{7}$
B.$\frac{2}{7}$
C.$\sqrt{2}$
D.$\frac{\sqrt{2}}{7}$
A
)A.$\frac{2\sqrt{5}}{7}$
B.$\frac{2}{7}$
C.$\sqrt{2}$
D.$\frac{\sqrt{2}}{7}$
答案
A
解析
$\begin{aligned}&\sqrt{1\frac{1}{3}} ÷ \sqrt{2\frac{1}{3}} ÷ \sqrt{1\frac{2}{5}}\\=&\sqrt{\frac{4}{3}} ÷ \sqrt{\frac{7}{3}} ÷ \sqrt{\frac{7}{5}}\\=&\sqrt{\frac{4}{3} ÷ \frac{7}{3} ÷ \frac{7}{5}}\\=&\sqrt{\frac{4}{3} × \frac{3}{7} × \frac{5}{7}}\\=&\sqrt{\frac{4×5}{7×7}}\\=&\frac{\sqrt{20}}{7}\\=&\frac{2\sqrt{5}}{7}\end{aligned}$
4. 化简$\sqrt{5} × \sqrt{\frac{9}{20}}$的结果是
$\frac{3}{2}$（或 1.5）
。答案
$\frac{3}{2}$（或 1.5）
解析
根据二次根式的乘法法则，$\sqrt{a} × \sqrt{b} = \sqrt{a × b}$，
有$\sqrt{5} × \sqrt{\frac{9}{20}} = \sqrt{5 × \frac{9}{20}} = \sqrt{\frac{45}{20}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$。
有$\sqrt{5} × \sqrt{\frac{9}{20}} = \sqrt{5 × \frac{9}{20}} = \sqrt{\frac{45}{20}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$。
5. 计算$\frac{\sqrt{5} × \sqrt{12}}{\sqrt{3}}$的结果是
$2\sqrt{5}$
。答案
$2\sqrt{5}$
解析
$\frac{\sqrt{5} × \sqrt{12}}{\sqrt{3}}=\sqrt{5}×\sqrt{\frac{12}{3}}=\sqrt{5}×\sqrt{4}=\sqrt{5}×2=2\sqrt{5}$
6. 计算：
(1)$\sqrt{15} × \sqrt{5}$；
(2)$2\sqrt{2} × 6\sqrt{6}$；
(3)$\sqrt{36} ÷ \sqrt{8}$；
(4)$\sqrt{2\frac{2}{3}} ÷ \sqrt{1\frac{3}{5}}$。
(1)$\sqrt{15} × \sqrt{5}$；
(2)$2\sqrt{2} × 6\sqrt{6}$；
(3)$\sqrt{36} ÷ \sqrt{8}$；
(4)$\sqrt{2\frac{2}{3}} ÷ \sqrt{1\frac{3}{5}}$。
答案
(1) $\sqrt{15} × \sqrt{5} = \sqrt{15×5} = \sqrt{75} = 5\sqrt{3}$
(2) $2\sqrt{2} × 6\sqrt{6} = (2×6)×(\sqrt{2}×\sqrt{6}) = 12×\sqrt{12} = 12×2\sqrt{3} = 24\sqrt{3}$
(3) $\sqrt{36} ÷ \sqrt{8} = \sqrt{36÷8} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$
(4) $\sqrt{2\frac{2}{3}} ÷ \sqrt{1\frac{3}{5}} = \sqrt{\frac{8}{3}} ÷ \sqrt{\frac{8}{5}} = \sqrt{\frac{8}{3}÷\frac{8}{5}} = \sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3}$
(2) $2\sqrt{2} × 6\sqrt{6} = (2×6)×(\sqrt{2}×\sqrt{6}) = 12×\sqrt{12} = 12×2\sqrt{3} = 24\sqrt{3}$
(3) $\sqrt{36} ÷ \sqrt{8} = \sqrt{36÷8} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$
(4) $\sqrt{2\frac{2}{3}} ÷ \sqrt{1\frac{3}{5}} = \sqrt{\frac{8}{3}} ÷ \sqrt{\frac{8}{5}} = \sqrt{\frac{8}{3}÷\frac{8}{5}} = \sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3}$
7. 计算：
(1)$\sqrt{8} × \sqrt{18}$；
(2)$\sqrt{2} × \sqrt{5} × \sqrt{10}$；
(3)$\sqrt{72} ÷ \frac{\sqrt{3}}{2}$；
(4)$-\sqrt{1\frac{2}{3}} ÷ \sqrt{\frac{5}{54}}$。
(1)$\sqrt{8} × \sqrt{18}$；
(2)$\sqrt{2} × \sqrt{5} × \sqrt{10}$；
(3)$\sqrt{72} ÷ \frac{\sqrt{3}}{2}$；
(4)$-\sqrt{1\frac{2}{3}} ÷ \sqrt{\frac{5}{54}}$。
答案
(1) $\sqrt{8} × \sqrt{18} = \sqrt{8×18} = \sqrt{144} = 12$；
(2) $\sqrt{2} × \sqrt{5} × \sqrt{10} = \sqrt{2×5×10} = \sqrt{100} = 10$；
(3) $\sqrt{72} ÷ \frac{\sqrt{3}}{2} = \sqrt{72} × \frac{2}{\sqrt{3}} = 2\sqrt{\frac{72}{3}} = 2\sqrt{24} = 2×2\sqrt{6} = 4\sqrt{6}$；
(4) $-\sqrt{1\frac{2}{3}} ÷ \sqrt{\frac{5}{54}} = -\sqrt{\frac{5}{3} ÷ \frac{5}{54}} = -\sqrt{\frac{5}{3}×\frac{54}{5}} = -\sqrt{18} = -3\sqrt{2}$。
(2) $\sqrt{2} × \sqrt{5} × \sqrt{10} = \sqrt{2×5×10} = \sqrt{100} = 10$；
(3) $\sqrt{72} ÷ \frac{\sqrt{3}}{2} = \sqrt{72} × \frac{2}{\sqrt{3}} = 2\sqrt{\frac{72}{3}} = 2\sqrt{24} = 2×2\sqrt{6} = 4\sqrt{6}$；
(4) $-\sqrt{1\frac{2}{3}} ÷ \sqrt{\frac{5}{54}} = -\sqrt{\frac{5}{3} ÷ \frac{5}{54}} = -\sqrt{\frac{5}{3}×\frac{54}{5}} = -\sqrt{18} = -3\sqrt{2}$。
8. 计算：
(1)$2\sqrt{12} × \frac{\sqrt{3}}{4} ÷ 10\sqrt{2}$；
(2)$\sqrt{7} ÷ \sqrt{3} × 2\sqrt{3} ÷ 2\sqrt{7}$；
(3)$\sqrt{10} × \sqrt{3} ÷ 2\sqrt{10} ÷ \frac{1}{6}\sqrt{3}$。
(1)$2\sqrt{12} × \frac{\sqrt{3}}{4} ÷ 10\sqrt{2}$；
(2)$\sqrt{7} ÷ \sqrt{3} × 2\sqrt{3} ÷ 2\sqrt{7}$；
(3)$\sqrt{10} × \sqrt{3} ÷ 2\sqrt{10} ÷ \frac{1}{6}\sqrt{3}$。
答案
(1)原式$=2× 2\sqrt{3}× \frac{\sqrt{3}}{4}÷ 10\sqrt{2}$
$=4\sqrt{3}× \frac{\sqrt{3}}{4}÷ 10\sqrt{2}$
$=3÷ 10\sqrt{2}$
$=\frac{3}{10\sqrt{2}}$
$=\frac{3\sqrt{2}}{20}$
(2)原式$=\sqrt{7}× \frac{1}{\sqrt{3}}× 2\sqrt{3}× \frac{1}{2\sqrt{7}}$
$=(\sqrt{7}× \frac{1}{2\sqrt{7}})× (\frac{1}{\sqrt{3}}× 2\sqrt{3})$
$=\frac{1}{2}× 2$
$=1$
(3)原式$=\sqrt{10}× \sqrt{3}× \frac{1}{2\sqrt{10}}× \frac{6}{\sqrt{3}}$
$=(\sqrt{10}× \frac{1}{2\sqrt{10}})× (\sqrt{3}× \frac{6}{\sqrt{3}})$
$=\frac{1}{2}× 6$
$=3$
$=4\sqrt{3}× \frac{\sqrt{3}}{4}÷ 10\sqrt{2}$
$=3÷ 10\sqrt{2}$
$=\frac{3}{10\sqrt{2}}$
$=\frac{3\sqrt{2}}{20}$
(2)原式$=\sqrt{7}× \frac{1}{\sqrt{3}}× 2\sqrt{3}× \frac{1}{2\sqrt{7}}$
$=(\sqrt{7}× \frac{1}{2\sqrt{7}})× (\frac{1}{\sqrt{3}}× 2\sqrt{3})$
$=\frac{1}{2}× 2$
$=1$
(3)原式$=\sqrt{10}× \sqrt{3}× \frac{1}{2\sqrt{10}}× \frac{6}{\sqrt{3}}$
$=(\sqrt{10}× \frac{1}{2\sqrt{10}})× (\sqrt{3}× \frac{6}{\sqrt{3}})$
$=\frac{1}{2}× 6$
$=3$
