《2025年暑期升级训练七年级数学浙教版浙江教育出版社》

6. 下列算式中, 运用分配律带来简便的是 ()

A.$60 ÷\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right)$
B.$\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right) ÷ 60$
C.$\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right) ÷ \frac{1}{60}$
D.$\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{12}\right) × \frac{1}{60}$

答案： C

7. 计算 $\left(-\frac{1}{2}\right)+\frac{1}{4}+\left(-\frac{2}{5}\right)+\left(+\frac{3}{10}\right)$ 时, 下列所运用的运算律恰当的是 ()

A.$\left[\left(-\frac{1}{2}\right)+\frac{1}{4}\right]+\left[\left(-\frac{2}{5}\right)+\left(+\frac{3}{10}\right)\right]$
B.$\left[\frac{1}{4}+\left(-\frac{2}{5}\right)\right]+\left[\left(-\frac{1}{2}\right)+\left(+\frac{3}{10}\right)\right]$
C.$\left(-\frac{1}{2}\right)+\left[\frac{1}{4}+\left(-\frac{2}{5}\right)\right]+\left(+\frac{3}{10}\right)$
D.$\left[\frac{1}{4}+\left(+\frac{3}{10}\right)\right]+\left[\left(-\frac{1}{2}\right)+\left(-\frac{2}{5}\right)\right]$

答案： A

8. 指出下列变形中用到的运算律.
(1) $(-2)+9+(-3)+2= [(-2)+2]+9+(-3)$, 根据的是加法的____.
(2) $(-3) ×(-8+2-3)= (-3) ×(-8)-3 × 2+3 × 3$, 根据的运算律是____.

答案：
(1) 交换律和结合律
(2) 分配律

9. 计算:
(1) $\frac{1}{6}+\left(-\frac{2}{7}\right)+\left(-\frac{5}{6}\right)+\left(+\frac{5}{7}\right)= $____.
(2) $-2.5+(-3.26)+5.5+(+7.26)= $____.
(3) $(-25) ×(-87) ×(-4)= $____.
(4) $\left(-\frac{6}{7}\right) ×(-15) ×\left(-\frac{7}{6}\right) × \frac{2}{5}= $____.
(5) $-9 \frac{22}{23} ×(-46)= $____.
(6) $-65 \frac{13}{15} ÷ 13= $____.
(7) $-81 ÷ 16 ÷\left(-2 \frac{1}{4}\right) ÷\left(-2 \frac{1}{4}\right)= $____.
(8) $\left(-\frac{1}{6}+\frac{3}{4}-\frac{1}{12}\right) ×(-48)= $____.
(9) $11.8 × 3 \frac{3}{4}-(-11.8) × 1.7-11.8 × \frac{3}{4}-11.8 ×(-0.3)= $____.
(10) $\frac{28}{5} ÷(-2) ×\left(-\frac{5}{14}\right)= $____.

答案：
(1) -$\frac{5}{21}$
(2) 7
(3) -8700
(4) -6
(5) 458
(6) -5$\frac{1}{15}$
(7) -1
(8) -24
(9) 59
(10) 1

10. 计算:
(1) $-5 \frac{5}{6}-9 \frac{2}{3}+17 \frac{3}{4}-3 \frac{1}{2}$.
(2) $-3.14 × 35+6.28 ×(-23.3)-15.7 × 3.68$.
(3) $\left(\frac{7}{9}-\frac{5}{6}+\frac{3}{18}\right) ÷\left(-\frac{1}{18}\right)-6 × 1.05-3.95 × 6$.
(4) $-3^2-7 ×\left(\frac{2}{7}-3\right)+12 ÷\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)$.
(5) $3.59 ×\left(-\frac{4}{7}\right)+2.41 ×\left(-\frac{4}{7}\right)-6 ×\left(-\frac{4}{7}\right)$.
(6) $5 × 401 × \frac{302}{1599}+(1599+401) × \frac{89}{1599}$.

答案：
(1) 原式 = -5 - $\frac{5}{6}$ - 9 - $\frac{2}{3}$ + 17 + $\frac{3}{4}$ - 3 - $\frac{1}{2}$ = (-5 - 9 + 17 - 3) + (-$\frac{5}{6}$ - $\frac{2}{3}$ + $\frac{3}{4}$ - $\frac{1}{2}$) = 0 - $\frac{5}{4}$ = -$\frac{5}{4}$.
(2) 原式 = -3.14 × (35 + 46.6 + 18.4) = -3.14 × 100 = -314.
(3) 原式 = -14 + 15 - 3 - 6 × (1.05 + 3.95) = 1 - 3 - 30 = -32.
(4) 原式 = -9 - 2 + 21 + 12 ÷ ($\frac{6}{12}$ - $\frac{4}{12}$ - $\frac{3}{12}$) = 10 + 12 ÷ (-$\frac{1}{12}$) = -134.
(5) 原式 = (-$\frac{4}{7}$) × (3.59 + 2.41 - 6) = (-$\frac{4}{7}$) × (6 - 6) = (-$\frac{4}{7}$) × 0 = 0.
(6) 原式 = 5 × 401 × $\frac{302}{1599}$ + 1599 × $\frac{89}{1599}$ + 401 × $\frac{89}{1599}$ = $\frac{401}{1599}$ × (5 × 302 + 89) + 89 = $\frac{401}{1599}$ × (1510 + 89) + 89 = $\frac{401}{1599}$ × 1599 + 89 = 401 + 89 = 490.