本题目所在的杯赛真题试卷：
🏆2025年19届-时代杯-真题-6年级
六年级
时代杯
2025
中等
计算技巧
速算与巧算（凑整、提取公因数等）
题目内容
题244【填空题】
神机妙算 计算下面各题｡
(1) \( \frac {9}{10} \div\left (\frac {1}{4} \times\left (\frac {6}{5}-\frac {1}{3}\right)\right) \) =_________
(2) \( \frac{1}{45} × 3 \frac{1}{7}+\frac{22}{63} ÷ 3 \frac{3}{14}+\frac{34}{45} × \frac{46}{63} \) =_________
(3) \( \frac{3}{1 × 2}-\frac{5}{2 × 3}+\frac{7}{3 × 4}-\frac{9}{4 × 5}+\cdots+\frac{199}{99 × 100}-\frac{201}{100 × 101} \) =_________
(4) \( \frac{1}{1^{3}}+\frac{1+2}{1^{3}+2^{3}}+\frac{1+2+3}{1^{3}+2^{3}+3^{3}}+\cdots+\frac{1+2+3+\cdots+2025}{1^{3}+2^{3}+3^{3}+\cdots+2025^{3}} \) =_________
(1) \( \frac {9}{10} \div\left (\frac {1}{4} \times\left (\frac {6}{5}-\frac {1}{3}\right)\right) \) =_________
(2) \( \frac{1}{45} × 3 \frac{1}{7}+\frac{22}{63} ÷ 3 \frac{3}{14}+\frac{34}{45} × \frac{46}{63} \) =_________
(3) \( \frac{3}{1 × 2}-\frac{5}{2 × 3}+\frac{7}{3 × 4}-\frac{9}{4 × 5}+\cdots+\frac{199}{99 × 100}-\frac{201}{100 × 101} \) =_________
(4) \( \frac{1}{1^{3}}+\frac{1+2}{1^{3}+2^{3}}+\frac{1+2+3}{1^{3}+2^{3}+3^{3}}+\cdots+\frac{1+2+3+\cdots+2025}{1^{3}+2^{3}+3^{3}+\cdots+2025^{3}} \) =_________
参考答案
(1)
\( \frac {54}{13} \)
；
(2) \( \frac {46}{63} \) ；
(3) \( \frac {100}{101} \) ；
(4) \( \frac {2025}{1013} \) ；
(2) \( \frac {46}{63} \) ；
(3) \( \frac {100}{101} \) ；
(4) \( \frac {2025}{1013} \) ；
题目解析
(1) 原式 =
\( \frac {9}{10} \div\left (\frac {1}{4} × \frac {13}{15}\right)=\frac {9}{10} × \frac {60}{13}=\frac {54}{13} \)
；
(2) 原式 = \( \frac {22}{45} × \frac {1}{7}+\frac {22}{45} × \frac {2}{9}+\frac {34}{45} × \frac {46}{63}=\frac {22}{45} \times\left (\frac {1}{7}+\frac {2}{9}\right)+\frac {34}{45} × \frac {46}{63}=\frac {22}{45} × \frac {23}{63}+\frac {23}{63} × \frac {68}{45}=\frac {23}{63} \times\left (\frac {22}{45}+\frac {68}{45}\right)=\frac {46}{63} \) ；
(3) 原式 = \( \left (\frac {1}{1}+\frac {1}{2}\right)-\left (\frac {1}{2}+\frac {1}{3}\right)+\left (\frac {1}{3}+\frac {1}{4}\right)-\left (\frac {1}{4}+\frac {1}{5}\right)+\cdots+\left (\frac {1}{99}+\frac {1}{100}\right)-\left (\frac {1}{100}+\frac {1}{101}\right)=1-\frac {1}{101}=\frac {100}{101} \) ；
(4) \( \text {原式}=\frac {1}{1} + \frac {1 + 2}{(1 + 2)^2} + \frac {1 + 2 + 3}{(1 + 2 + 3)^2} +\cdots + \frac {1 + 2 + 3 + \cdots + 2025}{(1 + 2 + 3 + \cdots + 2025)^2}=\frac {2}{1×2} + \frac {2}{2×3} + \frac {2}{3×4} + \cdots + \frac {2}{2025×2026}=2×\left (1 - \frac {1}{2} + \frac {1}{2} - \frac {1}{3} + \frac {1}{3} - \frac {1}{4} + \cdots + \frac {1}{2025} - \frac {1}{2026}\right)=2×\frac {2025}{2026}=\frac {2025}{1013} \)
(2) 原式 = \( \frac {22}{45} × \frac {1}{7}+\frac {22}{45} × \frac {2}{9}+\frac {34}{45} × \frac {46}{63}=\frac {22}{45} \times\left (\frac {1}{7}+\frac {2}{9}\right)+\frac {34}{45} × \frac {46}{63}=\frac {22}{45} × \frac {23}{63}+\frac {23}{63} × \frac {68}{45}=\frac {23}{63} \times\left (\frac {22}{45}+\frac {68}{45}\right)=\frac {46}{63} \) ；
(3) 原式 = \( \left (\frac {1}{1}+\frac {1}{2}\right)-\left (\frac {1}{2}+\frac {1}{3}\right)+\left (\frac {1}{3}+\frac {1}{4}\right)-\left (\frac {1}{4}+\frac {1}{5}\right)+\cdots+\left (\frac {1}{99}+\frac {1}{100}\right)-\left (\frac {1}{100}+\frac {1}{101}\right)=1-\frac {1}{101}=\frac {100}{101} \) ；
(4) \( \text {原式}=\frac {1}{1} + \frac {1 + 2}{(1 + 2)^2} + \frac {1 + 2 + 3}{(1 + 2 + 3)^2} +\cdots + \frac {1 + 2 + 3 + \cdots + 2025}{(1 + 2 + 3 + \cdots + 2025)^2}=\frac {2}{1×2} + \frac {2}{2×3} + \frac {2}{3×4} + \cdots + \frac {2}{2025×2026}=2×\left (1 - \frac {1}{2} + \frac {1}{2} - \frac {1}{3} + \frac {1}{3} - \frac {1}{4} + \cdots + \frac {1}{2025} - \frac {1}{2026}\right)=2×\frac {2025}{2026}=\frac {2025}{1013} \)
视频解析
None