《2025年重点中学招生分班五年真题分类卷数学》

15. (2024 陕西 JDFZ 少年班)$(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)$

答案： 解:令$1+0.12+0.23=a,0.12+0.23=b,$
则$a-b=1,$
原式$=a×(b+0.34)-(a+0.34)×b$
$=ab+0.34a-ab-0.34b$
$=0.34×(a-b)$
$=0.34$

16. (2024 陕西 GXYZ 博雅班)$(\frac {1}{9}+\frac {1}{10}+\frac {1}{11}+\frac {1}{12})×(\frac {1}{10}+\frac {1}{11}+\frac {1}{12}+\frac {1}{13})-(\frac {1}{9}+\frac {1}{10}+\frac {1}{11}+\frac {1}{12}+\frac {1}{13})×(\frac {1}{10}+\frac {1}{11}+\frac {1}{12})$

答案： 解:设$\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=a,$
原式$=a(a-\frac{1}{9}+\frac{1}{13})-(a+\frac{1}{13})(a-\frac{1}{9})$
$=a(a-\frac{1}{9})+\frac{1}{13}a-a(a-\frac{1}{9})-\frac{1}{13}a+\frac{1}{13}×\frac{1}{9}$
$=\frac{1}{13}×\frac{1}{9}$
$=\frac{1}{117}$

17. (2023 陕西 GDFZ)$(1-\frac {1}{2}-\frac {1}{3}-… -\frac {1}{2022})×(\frac {1}{2}+\frac {1}{3}+… +\frac {1}{2023})-(1-\frac {1}{2}-\frac {1}{3}-… -\frac {1}{2023})×(\frac {1}{2}+\frac {1}{3}+… +\frac {1}{2022})$

答案： 解:令$\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2023}=a$,令$\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2022}=b,$
则$a-b=\frac{1}{2023},$
原式$=(1-b)×a-(1-a)×b$
$=a-ab-b+ab$
$=a-b$
$=\frac{1}{2023}$

18. (2023 陕西 JDFZ)$(\frac {1}{2}+\frac {2}{3}+… +\frac {9}{10})^{2}+(\frac {1}{2}+\frac {2}{3}+… +\frac {9}{10})×\frac {1}{2}-(1+\frac {1}{2}+\frac {2}{3}+… +\frac {9}{10})×(\frac {2}{3}+\frac {3}{4}+… +\frac {9}{10})$

答案： 解:令$\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots +\frac{9}{10}=a,$
原式$=a^{2}+a×\frac{1}{2}-(1+a)(a-\frac{1}{2})$
$=a^{2}+\frac{a}{2}-(a^{2}+a-\frac{a}{2}-\frac{1}{2})$
$=a^{2}+\frac{a}{2}-a^{2}-a+\frac{a}{2}+\frac{1}{2}$
$=\frac{1}{2}$

19. 新考法 (2024 陕西 GXYZ 博雅班)$\frac {1}{2+\frac {1}{3+\frac {1}{4+\frac {1}…{ +\frac {1}{2009}}}}+\frac {1}{1+\frac {1}{1+\frac {1}{3+\frac {1}{4+\frac {1}…{ +\frac {1}{2009}}}}}$

答案： 解:设$3+\frac{1}{4+\frac{1}{\cdots +\frac{1}{2009}}}=a,$
原式$=\frac{1}{2+\frac{1}{a}}+\frac{1}{1+\frac{1}{1+\frac{1}{a}}}$
$=\frac{1}{\frac{2a+1}{a}}+\frac{1}{1+\frac{1}{\frac{a+1}{a}}}$
$=\frac{a}{2a+1}+\frac{1}{\frac{2a+1}{a+1}}$
$=\frac{a}{2a+1}+\frac{a+1}{2a+1}$
$=1$