《2025年重点中学招生分班五年真题分类卷数学》

1. (2024陕西XA3Z)999.3-998.2+997.3-996.2+…+3.3-2.2+1.3-0.2

答案： 解:原式=(999.3-998.2)+(997.3-996.2)+…+(1.3-0.2)
=1.1+1.1+1.1+…+1.1
=1.1×500
=550

2. (2024陕西JDFZ)$1+\frac {1}{2}+\frac {2}{2}+\frac {1}{2}+\frac {1}{3}+\frac {2}{3}+\frac {3}{3}+\frac {2}{3}+\frac {1}{3}+... +\frac {1}{2010}+\frac {2}{2010}+\frac {3}{2010}+... +\frac {2}{2010}+\frac {1}{2010}$

答案： 解:原式=1+($\frac{1}{2}+\frac{2}{2}+\frac{1}{2}$)+($\frac{1}{3}+\frac{2}{3}+\frac{3}{3}+\frac{2}{3}+\frac{1}{3}$)+…+$\frac{1}{2010}+\frac{2}{2010}+…+\frac{2010}{2010}+…+\frac{2}{2010}+\frac{1}{2010}$
=1+2+3+…+2010
=(1+2010)×2010÷2
=2021055

3. (2021陕西JDFZ)$[(10-\frac {2}{55}×1)+(9-\frac {2}{55}×2)+(8-\frac {2}{55}×3)+... +(1-\frac {2}{55}×10)]×\frac {5}{11}$

答案： 解:原式=(10-$\frac{2}{55}$×1+9-$\frac{2}{55}$×2+8-$\frac{2}{55}$×3+…+1-$\frac{2}{55}$×10)×$\frac{5}{11}$
=(10+9+8+…+1-$\frac{2}{55}$×1-$\frac{2}{55}$×2-$\frac{2}{55}$×3-…-$\frac{2}{55}$×10)×$\frac{5}{11}$
=[(10+9+8+…+1)-($\frac{2}{55}$×1+$\frac{2}{55}$×2+$\frac{2}{55}$×3+…+$\frac{2}{55}$×10)]×$\frac{5}{11}$
=[(10+9+8+…+1)-$\frac{2}{55}$×(1+2+3+…+10)]×$\frac{5}{11}$
=(10+9+8+…+1)×(1-$\frac{2}{55}$)×$\frac{5}{11}$
=55×$\frac{53}{55}$×$\frac{5}{11}$
=$\frac{265}{11}$

4. (2024重庆8Z)$\frac {1-\frac {1}{2}+\frac {1}{3}-\frac {1}{4}+\frac {1}{5}-... +\frac {1}{19}-\frac {1}{20}}{\frac {1}{1+21}+\frac {1}{2+22}+... +\frac {1}{10+30}}$

答案： 解:分子=(1-$\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}$)+($\frac{1}{3}-\frac{1}{6}-\frac{1}{12}$)+($\frac{1}{5}-\frac{1}{10}-\frac{1}{20}$)+($\frac{1}{7}-\frac{1}{14}$)+($\frac{1}{9}-\frac{1}{18}$)+$\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+\frac{1}{17}+\frac{1}{19}$
=$\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+…+\frac{1}{19}+\frac{1}{20}$
分母=$\frac{1}{22}+\frac{1}{24}+…+\frac{1}{40}$=$\frac{1}{2}$×($\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+…+\frac{1}{19}+\frac{1}{20}$)
原式=$\frac{\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+…+\frac{1}{19}+\frac{1}{20}}{\frac{1}{2}×(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+…+\frac{1}{19}+\frac{1}{20})}$\n=2

5. (2020陕西GXYZ)$\frac {1}{1+1}+\frac {2}{1+2}+\frac {3}{1+3}+... +\frac {2019}{1+2019}+\frac {\frac {1}{2}}{1+\frac {1}{2}}+\frac {\frac {1}{3}}{1+\frac {1}{3}}+... +\frac {\frac {1}{2019}}{1+\frac {1}{2019}}$

答案： 解:原式=$\frac{1}{1+1}$+($\frac{2}{1+2}+\frac{\frac{1}{2}}{1+\frac{1}{2}}$)+($\frac{3}{1+3}+\frac{\frac{1}{3}}{1+\frac{1}{3}}$)+…+($\frac{2019}{1+2019}+\frac{\frac{1}{2019}}{1+\frac{1}{2019}}$)
=$\frac{1}{2}$+($\frac{2}{3}+\frac{1}{3}$)+($\frac{3}{4}+\frac{1}{4}$)+…+($\frac{2019}{2020}+\frac{1}{2020}$)
=$\frac{1}{2}$+$\underbrace{1+1+…+1}_{2018个}$
=$\frac{1}{2}$+2018
=2018$\frac{1}{2}$

6. (2023陕西JDFZ)$41\frac {1}{3}×\frac {3}{4}+51\frac {1}{4}×\frac {4}{5}+61\frac {1}{5}×\frac {5}{6}+71\frac {1}{6}×\frac {6}{7}$

答案： 解:原式=(40+$\frac{4}{3}$)×$\frac{3}{4}$+(50+$\frac{5}{4}$)×$\frac{4}{5}$+(60+$\frac{6}{5}$)×$\frac{5}{6}$+(70+$\frac{7}{6}$)×$\frac{6}{7}$
=40×$\frac{3}{4}$+$\frac{4}{3}$×$\frac{3}{4}$+50×$\frac{4}{5}$+$\frac{5}{4}$×$\frac{4}{5}$+60×$\frac{5}{6}$+$\frac{6}{5}$×$\frac{5}{6}$+70×$\frac{6}{7}$+$\frac{7}{6}$×$\frac{6}{7}$
=30+1+40+1+50+1+60+1
=184