第131页 - 经纶学典学霸八年级数学上下册【江苏国标】

C

6

$2 \sqrt{15} $

$解:原式= \sqrt{\frac{3}{4}×\frac{8}{3}}=\sqrt{2}$

$解:原式= \sqrt{xy^{5}×x^{3}y×\frac{1}{y^{6}}}=x^{2}$

$解:原式=- \sqrt{108a^{3}b}×\sqrt {6}b=-18ab\sqrt {2a}$

$解:原式=\sqrt {(2025-2023)(2025+2023)}$

$=\sqrt{2×4048}$

$=4\sqrt{506}$

$解:原式=(2\sqrt {3})^{2}-(3\sqrt {2})^{2}$

$=12-18=-6$

$6 \sqrt{\frac{1}{7}}$

$24\sqrt {10}$

$解:(2)\sqrt {n+\frac {1}{n+2}}=(n+1) \sqrt{\frac{1}{n+2}}$

$\ 验证: \sqrt {n+\frac {1}{n+2}}=\sqrt {\frac {(n+1)^{2}}{n+2}}=(n+1)\sqrt {\frac {1}{n+2}}$

$(3)\sqrt{2022+\frac {1}{2024}}× \sqrt{4048}$

$=2023×\sqrt{\frac{1}{2024}}× \sqrt{4048}$

$=2023\sqrt {2}$