《2025年名校题库八年级数学上册北师大版》

7.(武侯区期末)计算:
(1)$\sqrt {24}-3\sqrt {\frac {2}{3}}-\sqrt [3]{-27};$
(2)$9\sqrt {8}÷3\sqrt {\frac {1}{5}}×2\sqrt {5}.$

答案：
(1)原式$=2\sqrt{6}-3×\frac{\sqrt{6}}{3}-(-3)$
$=2\sqrt{6}-\sqrt{6}+3$
$=\sqrt{6}+3$
(2)原式$=9×2\sqrt{2}÷\left(3×\frac{\sqrt{5}}{5}\right)×2\sqrt{5}$
$=18\sqrt{2}÷\frac{3\sqrt{5}}{5}×2\sqrt{5}$
$=18\sqrt{2}×\frac{5}{3\sqrt{5}}×2\sqrt{5}$
$=30\sqrt{2}×\frac{2\sqrt{5}}{\sqrt{5}}$
$=60\sqrt{2}$

8.(锦江区期末)计算:
(1)$(\frac {1}{3})^{2}+(π-2022)^{0}-\sqrt {9}+|2-\sqrt {5}|;$
(2)$\sqrt {24}÷\sqrt {1\frac {1}{2}}-\sqrt {\frac {1}{3}}×\sqrt {12}+\sqrt {48}.$

答案：
(1)解：原式$=\frac{1}{9}+1-3+\sqrt{5}-2=-\frac{35}{9}+\sqrt{5}$
(2)解：原式$=\sqrt{24×\frac{2}{3}}-\sqrt{\frac{1}{3}×12}+4\sqrt{3}=4-2+4\sqrt{3}=2+4\sqrt{3}$

9.(高新区期末)计算:
(1)$3\sqrt {27}+\frac {1}{5}\sqrt {75}-6\sqrt {\frac {1}{3}};$
(2)$\sqrt {18}-\frac {2}{\sqrt {2}}+(1-\sqrt {2})(1+\sqrt {2}).$

答案：
(1)解：原式$=3×3\sqrt{3}+\frac{1}{5}×5\sqrt{3}-6×\frac{\sqrt{3}}{3}$
$=9\sqrt{3}+\sqrt{3}-2\sqrt{3}$
$=8\sqrt{3}$
(2)解：原式$=3\sqrt{2}-\frac{2\sqrt{2}}{\sqrt{2}×\sqrt{2}}+(1^{2}-(\sqrt{2})^{2})$
$=3\sqrt{2}-\sqrt{2}+(1 - 2)$
$=2\sqrt{2}-1$

10.(成外)计算:
(1)$\sqrt {18}×\sqrt {\frac {2}{3}}-(1-\sqrt {3})^{2};$
(2)$(π-2018)^{0}+6\sqrt {\frac {1}{3}}-|5-\sqrt {27}|-(\frac {1}{2})^{-2};$
(3)$(-\frac {1}{2})^{-2}+\sqrt {(1-\sqrt {3})^{2}}-\frac {\sqrt {20}-\sqrt {15}}{\sqrt {5}}-|\sqrt {3}-1|.$

答案：
(1)解：原式$=\sqrt{18×\frac{2}{3}}-(1-2\sqrt{3}+3)$
$=\sqrt{12}-(4-2\sqrt{3})$
$=2\sqrt{3}-4+2\sqrt{3}$
$=4\sqrt{3}-4$
(2)解：原式$=1+6×\frac{\sqrt{3}}{3}-(3\sqrt{3}-5)-4$
$=1+2\sqrt{3}-3\sqrt{3}+5-4$
$=2-\sqrt{3}$
(3)解：原式$=4+(\sqrt{3}-1)-(\sqrt{4}-\sqrt{3})-(\sqrt{3}-1)$
$=4+\sqrt{3}-1-2+\sqrt{3}-\sqrt{3}+1$
$=2+\sqrt{3}$

11.(金牛区期末)计算:
(1)$\sqrt {8}+|1-\sqrt {2}|+(\frac {1}{2})^{-1}-2017^{0};$
(2)$\sqrt {27}×\sqrt {\frac {2}{3}}-(\sqrt {2}-1)^{2}.$

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