《2025年名校题库八年级数学上册北师大版》

12.(成华区期末)计算:
(1)$\sqrt {45}÷3\sqrt {3}×\sqrt {\frac {3}{5}};$
(2)$\sqrt [3]{8}+\frac {1}{2+\sqrt {5}}-(\frac {1}{3})^{2}+|\sqrt {5}-3|.$

答案：
(1)原式$=3\sqrt{5}÷3\sqrt{3}×\frac{\sqrt{15}}{5}$
$=\frac{\sqrt{5}}{\sqrt{3}}×\frac{\sqrt{15}}{5}$
$=\frac{\sqrt{5}×\sqrt{15}}{5\sqrt{3}}$
$=\frac{\sqrt{75}}{5\sqrt{3}}$
$=\frac{5\sqrt{3}}{5\sqrt{3}}$
$=1$
(2)原式$=2+\frac{2-\sqrt{5}}{(2+\sqrt{5})(2-\sqrt{5})}-\frac{1}{9}+3-\sqrt{5}$
$=2+\frac{2-\sqrt{5}}{4 - 5}-\frac{1}{9}+3-\sqrt{5}$
$=2+\frac{2 - \sqrt{5}}{-1}-\frac{1}{9}+3 - \sqrt{5}$
$=2 - 2 + \sqrt{5}-\frac{1}{9}+3 - \sqrt{5}$
$=(2 - 2 + 3) + (\sqrt{5} - \sqrt{5})-\frac{1}{9}$
$=3 - \frac{1}{9}$
$=\frac{27}{9}-\frac{1}{9}$
$=\frac{26}{9}$
$=2\frac{8}{9}$

13.(嘉祥)计算.
(1)$\sqrt {27}+\sqrt [3]{-8}+\sqrt {\frac {1}{3}}-\sqrt {12};$
(2)$2\sqrt {3}×\frac {\sqrt {2}}{4}-\sqrt {48}÷\sqrt {3}+(2-\sqrt {6})(2+\sqrt {6}).$

答案：
(1)解：原式$=3\sqrt{3}+(-2)+\frac{\sqrt{3}}{3}-2\sqrt{3}$
$=-2+\left(3\sqrt{3}-\frac{\sqrt{3}}{3}-2\sqrt{3}\right)$
$=-2+\frac{4\sqrt{3}}{3}$
(2)解：原式$=\frac{2\sqrt{3}×\sqrt{2}}{4}-\sqrt{48÷3}+(2^2-(\sqrt{6})^2)$
$=\frac{\sqrt{6}}{2}-\sqrt{16}+(4 - 6)$
$=\frac{\sqrt{6}}{2}-4-2$
$=\frac{\sqrt{6}}{2}-6$

14.(石室联中)计算:
(1)$\sqrt {8}-\sqrt [3]{27}+(\frac {\sqrt {2}}{2})^{-1};$
(2)$|1-\sqrt {2}|+(π-3)^{0}-6\sqrt {\frac {1}{2}};$
(3)$(2-\sqrt {5})(2+\sqrt {5})+(2-\sqrt {2})^{2}-\frac {1}{\sqrt {2}}.$

答案：
(1)解：原式$=2\sqrt{2}-3+\sqrt{2}=3\sqrt{2}-3$
(2)解：原式$=\sqrt{2}-1+1-3\sqrt{2}=-2\sqrt{2}$
(3)解：原式$=4-5+4-4\sqrt{2}+2-\frac{\sqrt{2}}{2}=5-\frac{9}{2}\sqrt{2}$

15.(树德实验)计算:
(1)$\sqrt {81}+(π-\sqrt {3})^{0}-\sqrt {5}+|2-\sqrt {5}|;$
(2)$-1^{2024}+\sqrt {(-2)^{2}}+|\sqrt {3}-2|+\sqrt [3]{-27};$
(3)$(3\sqrt {12}-2\sqrt {6}+\sqrt {48})÷2\sqrt {3}.$

答案：
(1)解：原式$=9 + 1 - \sqrt{5} + (\sqrt{5} - 2)$
$=9 + 1 - \sqrt{5} + \sqrt{5} - 2$
$=8$
(2)解：原式$=-1 + 2 + (2 - \sqrt{3}) + (-3)$
$=-1 + 2 + 2 - \sqrt{3} - 3$
$=-\sqrt{3}$
(3)解：原式$=3\sqrt{12}÷2\sqrt{3} - 2\sqrt{6}÷2\sqrt{3} + \sqrt{48}÷2\sqrt{3}$
$=3×2\sqrt{3}÷2\sqrt{3} - 2\sqrt{6}÷2\sqrt{3} + 4\sqrt{3}÷2\sqrt{3}$
$=3 - \sqrt{2} + 2$
$=5 - \sqrt{2}$