2026年通城学典初中数学运算能手八年级数学上册北师大版第88页答案
一、填空题
1. $(\dfrac{1}{3})^0 - (\dfrac{1}{4})^{-1} + \sqrt{(-2)^2} =$
-1

2. $\sqrt{25} - \sqrt[3]{8} + \sqrt{\dfrac{4}{9}} =$
$\dfrac{11}{3}$

3. $-5\sqrt{3} + 4\sqrt{3} + \dfrac{\sqrt{3}}{3} =$
$-\dfrac{2\sqrt{3}}{3}$

4. $|2+\sqrt{3}| + |\sqrt{3}-2| =$
$4$

5. $|2\sqrt{2} -4| - \sqrt[3]{27} + 3\sqrt{2} =$
$1+\sqrt{2}$

6. $\sqrt[3]{125} - \sqrt{9} + |\sqrt{5}-2| =$
$\sqrt{5}$

7. $\sqrt{(-\dfrac{1}{3})^2} - \sqrt{\dfrac{25}{81}} + \sqrt[3]{-64} =$
$-4\dfrac{2}{9}$

8. $\sqrt{32} + \sqrt[3]{8} - |π^0 - \sqrt{2}| - (\dfrac{1}{3})^{-1} =$
$3\sqrt{2}$

答案

1. -1
2. $\dfrac{11}{3}$
3. $-\dfrac{2\sqrt{3}}{3}$
4. 4
5. $1+\sqrt{2}$
6. $\sqrt{5}$
7. $-4\dfrac{2}{9}$
8. $3\sqrt{2}$
二、计算题
9. $\sqrt[3]{-27}+\sqrt{(-3)^2}-\sqrt[3]{-1}$
10. $\sqrt{6}+2\sqrt{3}-5(\sqrt{6}+2\sqrt{3})$
11. $\sqrt[3]{-216}+\sqrt{16}×\sqrt{\frac{9}{4}}÷(-\sqrt{3})^2$
12. $2\sqrt{7}-\sqrt{2}-|\sqrt{7}-\sqrt{2}|$
13. $3(\sqrt{2}+\sqrt{3})-2(\sqrt{2}-\sqrt{3})$
14. $(-4^2)÷\sqrt{\frac{16}{81}}×(-\frac{3}{2})^2+81$
15. 一题多解 $(-63)×(\sqrt{2\frac{7}{9}}+\frac{5}{9}-\sqrt[3]{\frac{1}{343}})$
16. $(3.14-π)^0+|\sqrt{2}-1|+(\frac{1}{2})^{-1}-\sqrt{8}$

答案

9. 1
10. $-4\sqrt{6}-8\sqrt{3}$
11. -4
12. $\sqrt{7}$
13. $\sqrt{2}+5\sqrt{3}$
14. 0
15. 解法一 原式$=(-63)×(\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{1}{7})=(-63)× \dfrac{131}{63}=-131.$
解法二 原式$=(-63)×(\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{1}{7})=(-63)×\dfrac{5}{3}+(-63)×\dfrac{5}{9}+(-63)×(-\dfrac{1}{7})=-105-35+9=-131.$
16. $2-\sqrt{2}$