《2025年暑期升级训练七年级数学浙教版浙江教育出版社》

11. 计算: $\frac{1}{24} ÷\left(\frac{2}{3}-\frac{3}{4}+\frac{1}{6}-\frac{5}{12}\right)$.

答案： 因为 ($\frac{2}{3}$ - $\frac{3}{4}$ + $\frac{1}{6}$ - $\frac{5}{12}$) ÷ $\frac{1}{24}$ = ($\frac{2}{3}$ - $\frac{3}{4}$ + $\frac{1}{6}$ - $\frac{5}{12}$) × 24 = $\frac{2}{3}$ × 24 - $\frac{3}{4}$ × 24 + $\frac{1}{6}$ × 24 - $\frac{5}{12}$ × 24 = -8, 所以 $\frac{1}{24}$ ÷ ($\frac{2}{3}$ - $\frac{3}{4}$ + $\frac{1}{6}$ - $\frac{5}{12}$) = -$\frac{1}{8}$.
方法点金
利用转化思想进行简便运算
除法没有分配律,若将被除数和除数交换位置,将除法转化为乘法,则可用分配律进行简便计算,此时结果与原式的结果互为倒数.

12. 计算: $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$.

答案： 原式 = $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ + $\frac{1}{4×5}$ + $\frac{1}{5×6}$ + $\frac{1}{6×7}$ + $\frac{1}{7×8}$ + $\frac{1}{8×9}$ + $\frac{1}{9×10}$ = (1 - $\frac{1}{2}$) + ($\frac{1}{2}$ - $\frac{1}{3}$) + ($\frac{1}{3}$ - $\frac{1}{4}$) + ($\frac{1}{4}$ - $\frac{1}{5}$) + ($\frac{1}{5}$ - $\frac{1}{6}$) + ($\frac{1}{6}$ - $\frac{1}{7}$) + ($\frac{1}{7}$ - $\frac{1}{8}$) + ($\frac{1}{8}$ - $\frac{1}{9}$) + ($\frac{1}{9}$ - $\frac{1}{10}$) = 1 - $\frac{1}{10}$ = $\frac{9}{10}$.

13. 阅读下面的计算方法:
$-5 \frac{5}{6}+\left(-9 \frac{2}{3}\right)+17 \frac{1}{2}+\left(-3 \frac{1}{2}\right)= \left[(-5)+\left(-\frac{5}{6}\right)\right]+\left[(-9)+\left(-\frac{2}{3}\right)\right]+\left(17+\frac{1}{2}\right)+\left[(-3)+\left(-\frac{1}{2}\right)\right]=[(-5)+(-9)+17+(-3)]+\left[\left(-\frac{5}{6}\right)+\left(-\frac{2}{3}\right)+\frac{1}{2}+\left(-\frac{1}{2}\right)\right]=0+\left(-1 \frac{1}{2}\right)= -1 \frac{1}{2}$.
(1) 计算: $\left(-2022 \frac{5}{6}\right)+\left(-2022 \frac{2}{3}\right)+\left(-1 \frac{1}{2}\right)+4045$.
(2) 已知 $1-\frac{1}{2^2}= \frac{1}{2} × \frac{3}{2}, 1-\frac{1}{3^2}= \frac{2}{3} × \frac{4}{3}, 1-\frac{1}{4^2}= \frac{3}{4} × \frac{5}{4}, …$. 计算: $\left(1-\frac{1}{2^2}\right) ×\left(1-\frac{1}{3^2}\right) × … ×\left(1-\frac{1}{2021^2}\right) ×\left(1-\frac{1}{2022^2}\right)$.

答案：
(1) 原式 = [(-2022) + (-$\frac{5}{6}$)] + [(-2022) + (-$\frac{2}{3}$)] + [(-1) + (-$\frac{1}{2}$)] + 4045 = [(-2022) + (-2022) + (-1) + 4045] + [(-$\frac{5}{6}$) + (-$\frac{2}{3}$) + (-$\frac{1}{2}$)] = 0 + (-2) = -2.
(2) 因为 1 - $\frac{1}{2^{2}}$ = $\frac{1}{2}$ × $\frac{3}{2}$, 1 - $\frac{1}{3^{2}}$ = $\frac{2}{3}$ × $\frac{4}{3}$, 1 - $\frac{1}{4^{2}}$ = $\frac{3}{4}$ × $\frac{5}{4}$, …, 所以原式 = $\frac{1}{2}$ × $\frac{3}{2}$ × $\frac{2}{3}$ × $\frac{4}{3}$ × $\frac{3}{4}$ × $\frac{5}{4}$ × … × $\frac{2020}{2021}$ × $\frac{2022}{2021}$ × $\frac{2021}{2022}$ × $\frac{2023}{2022}$ = $\frac{1}{2}$ × $\frac{2023}{2022}$ = $\frac{2023}{4044}$.