《2025年名校课堂八年级数学上册北师大版贵州专版》

1. 计算$\sqrt{27}-\sqrt{8}×\sqrt{\frac{3}{2}}$的结果是(A)

A.$\sqrt{3}$
B.$-\sqrt{3}$
C.$2\sqrt{3}$
D.$5\sqrt{3}$

答案： A

2. 老师设计了一个“接力游戏”，让同学们用合作的方式完成二次根式的混合运算.如图，老师把题目交给一位同学，他完成一步解答后交给第二位同学，依次进行，最后完成运算.规则是每人只能看到前一人传过来的式子.接力中，自己负责的运算出现错误的是(B)
$\begin{array}{c}\begin{array}{c}\boxed{\begin{array}{c}老师\\(\sqrt{12}+\sqrt{\frac{1}{18}})÷\sqrt{3}\end{array}}\end{array}\begin{array}{c}\Rightarrow\end{array}\begin{array}{c}\boxed{\begin{array}{c}小明\\\sqrt{12÷3}+\sqrt{\frac{1}{18}÷3}\end{array}}\end{array}\begin{array}{c}\Rightarrow\end{array}\begin{array}{c}\boxed{\begin{array}{c}小丽\\\sqrt{4}+\sqrt{\frac{1}{6}}\end{array}}\end{array}\\\begin{array}{c}\Rightarrow\end{array}\begin{array}{c}\boxed{\begin{array}{c}小红\\2+\sqrt{\frac{1}{36}}\end{array}}\end{array}\begin{array}{c}\Rightarrow\end{array}\begin{array}{c}\boxed{\begin{array}{c}小亮\\2\frac{1}{6}\end{array}}\end{array}\end{array}$

A.小明和小丽
B.小丽和小红
C.小红和小亮
D.小丽和小亮

答案： B

3. 在$(\sqrt{27}+\sqrt{\frac{1}{3}})□\sqrt{3}$的$□$中添加一个运算符号，使其结果最大，则$□$中应添加(C)

A.$+$
B.$-$
C.$×$
D.$÷$

答案： C

4. 计算：
(1) $\sqrt{24}-\sqrt{\frac{6}{5}}×\sqrt{45}=$$-\sqrt{6}$.
(2) $(-\sqrt{12}+\sqrt{\frac{1}{3}})÷\sqrt{27}=$$-\frac{5}{9}$.

答案：
(1)$-\sqrt{6}$
(2)$-\frac{5}{9}$

5. 计算：
(1) $\sqrt{\frac{2}{5}}-\sqrt{\frac{5}{2}}$.
(2) $\sqrt{80}-\sqrt{20}+\sqrt{\frac{1}{5}}$.
(3) $\frac{\sqrt{50}×\sqrt{32}}{\sqrt{8}}-\sqrt[3]{-8}$.
(4) $\sqrt{12}÷(\frac{\sqrt{3}}{4}+\frac{2\sqrt{3}}{3})$.

答案：
(1)原式$=\frac{\sqrt{10}}{5}-\frac{\sqrt{10}}{2}=-\frac{3\sqrt{10}}{10}$
(2)原式$=4\sqrt{5}-2\sqrt{5}+\frac{\sqrt{5}}{5}=2\sqrt{5}+\frac{1}{5}\sqrt{5}=\frac{11}{5}\sqrt{5}$.
(3)原式$=\frac{5\sqrt{2 × 4\sqrt{2}}}{2\sqrt{2}}-(-2)=10\sqrt{2}+2$.
(4)原式$=\sqrt{12} ÷\left(\frac{3\sqrt{3}}{12}+\frac{8\sqrt{3}}{12}\right)=\sqrt{12} ÷ \frac{11\sqrt{3}}{12}=2\sqrt{3} × \frac{12}{11\sqrt{3}}=\frac{24}{11}$.

6. 在计算$(\sqrt{24}-\sqrt{\frac{1}{2}})-(\sqrt{\frac{1}{8}}+\sqrt{6})$时，小敏的解题过程如下：
解：原式$=(2\sqrt{6}-\frac{\sqrt{2}}{2})-(\frac{\sqrt{2}}{4}+\sqrt{6})\cdots$①
$=2\sqrt{6}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}+\sqrt{6}\cdots$②
$=2\sqrt{6}+\sqrt{6}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}\cdots$③
$=3\sqrt{6}-\frac{\sqrt{2}}{4}\cdots$④
(1) 老师判定小敏的解题过程错误，请指出小敏是从第②步开始出错的.(填序号)
(2) 请写出正确的解题过程.

答案：
(1)②
(2)原式$=(2\sqrt{6}-\frac{\sqrt{2}}{2})-(\frac{\sqrt{2}}{4}+\sqrt{6})=2\sqrt{6}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}-\sqrt{6}=2\sqrt{6}-\sqrt{6}-(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{4})=\sqrt{6}-\frac{3\sqrt{2}}{4}$.