《2025年重点中学招生分班五年真题分类卷数学》

11. (2024陕西SDFZ)$\frac {4\frac {2}{7}+1\frac {1}{9}-0.28×7}{2\frac {1}{7}+\frac {5}{9}-0.49×2}÷\frac {1×2+2×3+... +2018×2019}{2×4+4×6+... +4036×4038}$

答案： 解:原式=$\frac{2×(\frac{15}{7}+\frac{5}{9}-0.98)}{\frac{15}{7}+\frac{5}{9}-0.98}÷\frac{1×2+2×3+\cdots+2018×2019}{4×(1×2+2×3+\cdots+2018×2019)}$
=2÷$\frac{1}{4}$
=8

12. (2024陕西XA1Z少年班)$\frac {(\frac {4}{7}×1\frac {1}{9}×\frac {4}{11})÷(\frac {2}{11}×\frac {2}{7}×\frac {5}{9})}{(\frac {4}{7}+1\frac {1}{9}+\frac {4}{11})÷(\frac {2}{11}+\frac {2}{7}+\frac {5}{9})}$

答案： 解:原式=$\frac{(\frac{4}{7}×\frac{10}{9}×\frac{4}{11})÷(\frac{2}{11}×\frac{2}{7}×\frac{5}{9})}{(\frac{4}{7}+\frac{10}{9}+\frac{4}{11})÷(\frac{2}{11}+\frac{2}{7}+\frac{5}{9})}$
=$\frac{(2×\frac{2}{7}×2×\frac{5}{9}×2×\frac{2}{11})÷(\frac{2}{11}×\frac{2}{7}×\frac{5}{9})}{(2×\frac{2}{7}+2×\frac{5}{9}+2×\frac{2}{11})÷(\frac{2}{11}+\frac{2}{7}+\frac{5}{9})}$
=$\frac{(2×2×2)×[(\frac{2}{7}×\frac{5}{9}×\frac{2}{11})÷(\frac{2}{11}×\frac{2}{7}×\frac{5}{9})]}{2×[(\frac{2}{7}+\frac{5}{9}+\frac{2}{11})÷(\frac{2}{11}+\frac{2}{7}+\frac{5}{9})]}$
=$\frac{2×2×2×1}{2×1}$
=4

13. (2024陕西JDFZ)$\frac {1^{2}+2^{2}+3^{2}+... +19^{2}+20^{2}}{2^{2}+4^{2}+6^{2}+... +38^{2}+40^{2}}$

答案： 解:原式=$\frac{1^{2}+2^{2}+3^{2}+\cdots+19^{2}+20^{2}}{(2×1)^{2}+(2×2)^{2}+(2×3)^{2}+\cdots+(2×19)^{2}+(2×20)^{2}}$
=$\frac{1^{2}+2^{2}+3^{2}+\cdots+19^{2}+20^{2}}{2^{2}×1^{2}+2^{2}×2^{2}+2^{2}×3^{2}+\cdots+2^{2}×19^{2}+2^{2}×20^{2}}$
=$\frac{1^{2}+2^{2}+3^{2}+\cdots+19^{2}+20^{2}}{2^{2}×(1^{2}+2^{2}+3^{2}+\cdots+19^{2}+20^{2})}$
=$\frac{1}{2^{2}}$
=$\frac{1}{4}$

14. (2022陕西JDFZ)$\frac {222×333+333×444}{555×777}$

答案： 解:原式=$\frac{333×(222+444)}{555×777}$
=$\frac{111×3×111×6}{111×5×111×7}$
=$\frac{18}{35}$

15. (2021陕西JDFZ)$\frac {12345654321}{777777×999999}$

答案： 解:原式=$\frac{111111×111111}{7×111111×9×111111}$
=$\frac{1}{63}$