2026年通城学典初中数学运算能手八年级上册苏科版第44页答案
一、填空题
1. $5+(-\sqrt{16}) × 3 = $
-7

2. $(-3)+(-\sqrt{64}) × (-2) = $
13

3. $[(-3)+\sqrt[3]{-64}] × (-2) = $
14

4. $|\sqrt{2}-\sqrt{3}|+\sqrt{(-5)^2}-\sqrt{3} = $
$5-\sqrt{2}$

5. $\sqrt{16} × [(-5)-(-25)] ÷ 5 = $
16

6. $(-5)-(-\sqrt{9}) ÷ \dfrac{1}{2}-15 ÷ \sqrt[3]{-27} = $
6

答案

1. -7
2. 13
3. 14
4. 5-√2
5. 16
6. 6
二、计算题
7. $(-15)-(-\sqrt{36}) ÷ \sqrt[3]{\frac{1}{27}} - 45 ÷ (-5)$
8. $(-32)-[\sqrt[3]{-8} ×(-8)-16 ÷(-8)] ÷(-\sqrt{36})$
9. $(-3)-(-\frac{3}{2})^2 × \sqrt{\frac{16}{81}} - \sqrt[3]{-216} ÷ \left|-\frac{2}{3}\right|^3$
10. $(-1)-[1-(1-\sqrt[3]{0.125} × \sqrt{\frac{1}{9}})-2^2] × 6$
11. $\sqrt[3]{-\frac{27}{64}} × 4^3 - \sqrt[3]{216} × \left|-\frac{2}{3}\right|^2 ÷ \sqrt{\frac{16}{25}}$
12. $\sqrt[3]{\frac{7}{8}-1} + \sqrt{\frac{1}{64}} - \sqrt[3]{1-\frac{189}{64}} - \sqrt{1-\frac{31}{256}}$
13. $-2^4+2 × \sqrt{5}+\sqrt[3]{8} ÷(\frac{1}{6}-\frac{1}{3}) ×[-\sqrt{(-\frac{1}{2})^2}]-|\sqrt{3}-2 × \sqrt{5}|$

答案

7. 12
8. -29
9. $16\dfrac{1}{4}$
10. 22
11. $-51\dfrac{1}{3}$
12. $-\dfrac{1}{16}$
13. $-10+\sqrt{3}$