《2025年重点中学招生分班五年真题分类卷数学》

7. (2023 陕西 TYZ)$25.5\% ÷[3-(5.55×1\frac {1}{3}-2\frac {1}{10}÷0.4)]$

答案： 解:原式=25.5%÷[3-($\frac{111}{20}$×$\frac{4}{3}$-$\frac{21}{10}$×$\frac{5}{2}$)]
=25.5%÷[3-($\frac{37}{5}$-$\frac{21}{4}$)]
=25.5%÷$\frac{17}{20}$
=0.3

8. (2023 陕西 JDFZ)$1.\dot {2}×1.\dot {2}\dot {4}+\frac {19}{27}$

答案： 解:原式=1$\frac{2}{9}$×1$\frac{8}{33}$+$\frac{19}{27}$
=$\frac{11}{9}$×$\frac{41}{33}$+$\frac{19}{27}$
=$\frac{41}{27}$+$\frac{19}{27}$
=$\frac{60}{27}$
=2$\frac{2}{9}$

9. (2022 陕西 TYZ)$3.672÷0.08÷\frac {2}{13}+3\frac {17}{50}×3.95÷0.1-33\frac {2}{5}÷2\frac {2}{9}$

答案： 解:原式=45.9×6.5+3.34×39.5-3.34×4.5
=298.35+3.34×(39.5-4.5)
=298.35+116.9
=415.25

10. (2023 陕西 TYZ)$15-3\frac {1}{8}÷(1\frac {2}{3}-0.625)×(1.6+2\frac {1}{3})$

答案： 解:原式=15-$\frac{25}{8}$÷(1$\frac{2}{3}$-$\frac{5}{8}$)×(1$\frac{3}{5}$+2$\frac{1}{3}$)
=15-$\frac{25}{8}$×$\frac{24}{25}$×$\frac{59}{15}$
=15-11$\frac{4}{5}$
=3$\frac{1}{5}$

11. (2024 陕西 TYZBH 学校)$\frac {1}{1+\frac {1}{2+\frac {1}{3+\frac {1}{4}}}}$

答案： 解:原式=$\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}$
=$\frac{1}{1+\frac{1}{2+\frac{4}{13}}}$
=$\frac{1}{1+\frac{13}{30}}$
=$\frac{30}{43}$