《2025年重点中学招生分班五年真题分类卷数学》

7. (2024河南ZZ47Z)$999\frac {8}{9}+99\frac {8}{9}+9\frac {8}{9}+\frac {1}{3}$

答案： 解:原式=999$\frac{8}{9}$+99$\frac{8}{9}$+9$\frac{8}{9}$+($\frac{1}{9}$×3)
=999+$\frac{8}{9}$+99+$\frac{8}{9}$+9+$\frac{8}{9}$+$\frac{1}{9}$+$\frac{1}{9}$+$\frac{1}{9}$
=(999+$\frac{8}{9}$+$\frac{1}{9}$)+(99+$\frac{8}{9}$+$\frac{1}{9}$)+(9+$\frac{8}{9}$+$\frac{1}{9}$)
=1000+100+10
=1110

8. (2024陕西TYZ)$1\frac {1}{2}+3\frac {1}{6}+5\frac {1}{12}+7\frac {1}{20}+9\frac {1}{30}$

答案： 解:原式=1+3+5+7+9+$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}$
=25+(1-$\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}$)
=25$\frac{5}{6}$

9. (2024陕西TYZ)$2011×\frac {2013}{2012}+2012×\frac {2014}{2013}+\frac {4025}{2012×2013}$

答案： 解:原式=$\frac{(2012-1)(2012+1)}{2012}$+$\frac{(2013-1)(2013+1)}{2013}$+$\frac{2012+2013}{2012×2013}$
=$\frac{2012^{2}-1}{2012}$+$\frac{2013^{2}-1}{2013}$+$\frac{1}{2012}$+$\frac{1}{2013}$
=2012-$\frac{1}{2012}$+2013-$\frac{1}{2013}$+$\frac{1}{2012}$+$\frac{1}{2013}$
=2012+2013
=4025

10. (2023陕西TYZ)123456+234561+345612+456123+561234+612345

答案： 解:原式=(1×100000+2×10000+3×1000+4×100+5×10+6×1)+(2×100000+3×10000+4×1000+5×100+6×10+1×1)+…+(6×100000+1×10000+2×1000+3×100+4×10+5×1)
=111111+222222+333333+444444+555555+666666
=(1+2+3+4+5+6)×111111
=21×111111
=(20+1)×111111
=2222220+111111
=2333331

11. (2023陕西GDFZ)$\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+\frac {1}{30}+\frac {1}{42}$

答案： 解:原式=$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+\frac{1}{6×7}$
=1-$\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}$
=1-$\frac{1}{7}$
=$\frac{6}{7}$

12. (2024陕西XA83Z)$\frac {4}{2×4}+\frac {4}{4×6}+\frac {4}{6×8}+\frac {4}{8×10}+\frac {4}{10×12}+... +\frac {4}{360}$

答案： 解:原式=$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+…+\frac{1}{90}$
=$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}+\frac{1}{5×6}+…+\frac{1}{9×10}$
=1-$\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{9}-\frac{1}{10}$
=1-$\frac{1}{10}$
=$\frac{9}{10}$

13. (2022陕西TYZ)$6\frac {1}{3}+12\frac {1}{6}+20\frac {1}{10}+30\frac {1}{15}+42\frac {1}{21}+56\frac {1}{28}+72\frac {1}{36}$

答案： 解:原式=(6+12+20+30+42+56+72)+($\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}$)
=238+[$\frac{1}{3}+\frac{1}{6}+\frac{1}{3}×(\frac{1}{2}-\frac{1}{5})+\frac{1}{15}+\frac{1}{4}×(\frac{1}{3}-\frac{1}{7})+\frac{1}{28}+\frac{1}{36}$]
=238+($\frac{1}{3}+\frac{1}{6}+\frac{1}{6}-\frac{1}{15}+\frac{1}{15}+\frac{1}{12}-\frac{1}{28}+\frac{1}{28}+\frac{1}{36}$)
=238+($\frac{1}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{12}+\frac{1}{36}$)
=238$\frac{7}{9}$

14. ( 2024陕西JDFZ)$\frac {1}{3×33}+\frac {3×33+1}{33×333}+\frac {3×333+1}{333×3333}+\frac {3×3333+1}{3333×33333}$

答案： 解:原式=$\frac{1}{30}×(\frac{1}{3}-\frac{1}{33})+\frac{1}{30}×(\frac{1}{33}-\frac{1}{333})+\frac{1}{30}×(\frac{1}{333}-\frac{1}{3333})+\frac{1}{30}×(\frac{1}{3333}-\frac{1}{33333})$
=$\frac{1}{30}×(\frac{1}{3}-\frac{1}{33}+\frac{1}{33}-\frac{1}{333}+\frac{1}{333}-\frac{1}{3333}+\frac{1}{3333}-\frac{1}{33333})$
=$\frac{1}{30}×(\frac{1}{3}-\frac{1}{33333})$
=$\frac{1}{30}×\frac{11110}{33333}$
=$\frac{1111}{99999}$

15. (2024陕西TYZ)$\frac {3^{2}+1}{3^{2}-1}+\frac {5^{2}+1}{5^{2}-1}+\frac {7^{2}+1}{7^{2}-1}+\frac {9^{2}+1}{9^{2}-1}+... +\frac {99^{2}+1}{99^{2}-1}$

答案： 解:原式=$\frac{9+1}{(3+1)(3-1)}+\frac{25+1}{(5+1)(5-1)}+\frac{49+1}{(7+1)(7-1)}+\frac{81+1}{(9+1)(9-1)}+…+\frac{9801+1}{(99+1)(99-1)}$
=$\frac{10}{2×4}+\frac{26}{4×6}+\frac{50}{6×8}+\frac{82}{8×10}+…+\frac{9802}{98×100}$
=1+$\frac{2}{2×4}+1+\frac{2}{4×6}+1+\frac{2}{6×8}+1+\frac{2}{8×10}+…+1+\frac{2}{98×100}$
=49+($\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+…+\frac{1}{98}-\frac{1}{100}$)
=49+($\frac{1}{2}-\frac{1}{100}$)
=49+$\frac{49}{100}$
=49$\frac{49}{100}$