《2025年重点中学招生分班五年真题分类卷数学》

6. (2024陕西GDFZ)$(1\frac {5}{88}+2\frac {5}{44}+8\frac {5}{11})÷(1\frac {1}{88}+2\frac {1}{44}+8\frac {1}{11})$

答案： 解:原式=$(\frac{93}{88}+\frac{93}{44}+\frac{93}{11})÷(\frac{89}{88}+\frac{89}{44}+\frac{89}{11})$
=$[93×(\frac{1}{88}+\frac{1}{44}+\frac{1}{11})]÷[89×(\frac{1}{88}+\frac{1}{44}+\frac{1}{11})]$
=93÷89
=$\frac{93}{89}$

7. (2024陕西JDKG学校)$\frac {2012+2014×2011}{2012×2014-2}+\frac {9\frac {2}{7}+7\frac {2}{9}}{\frac {5}{7}+\frac {5}{9}}$

答案： 解:原式=$\frac{2012+2014×(2012-1)}{2012×2014-2}+\frac{\frac{65}{7}+\frac{65}{9}}{\frac{5}{7}+\frac{5}{9}}$
=$\frac{2012×2014-2}{2012×2014-2}+\frac{65×(\frac{1}{7}+\frac{1}{9})}{5×(\frac{1}{7}+\frac{1}{9})}$
=$\frac{2012×2014-2}{2012×2014-2}+\frac{65}{5}$
=1+13
=14

8. (2023陕西GXYZSY中学)$\frac {\frac {6+7+8}{5}-\frac {7+8+9}{6}+\frac {8+9+10}{7}-\frac {9+10+11}{8}}{\frac {1}{5}-\frac {1}{6}+\frac {1}{7}-\frac {1}{8}}$

答案： 解:原式=$\frac{(3+\frac{6}{5})-(3+\frac{6}{6})+(3+\frac{6}{7})-(3+\frac{6}{8})}{\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}}$
=$\frac{\frac{6}{5}-\frac{6}{6}+\frac{6}{7}-\frac{6}{8}}{\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}}$
=$\frac{6×(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8})}{\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}}$
=6

9. (2022陕西GXYZ博雅班)$(\frac {5\frac {1}{3}+2\frac {1}{4}-1\frac {1}{5}}{1\frac {1}{3}+150\% }+1)×\frac {1\frac {1}{3}+150\% }{5\frac {1}{3}+(1\frac {1}{3}+150\% )+2\frac {1}{4}-1\frac {1}{5}}$

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

10. (2024陕西GX3C)$\frac {4×5×6+8×10×12+16×20×24}{2×3×4+4×6×8+8×12×16}$

答案： 解:原式=$\frac{4×5×6+4×2×5×2×6×2+4×4×5×4×6×4}{2×3×4+2×2×3×2×4×2+2×4×3×4×4×4}$
=$\frac{4×5×6+4×5×6×2^{3}+4×5×6×4^{3}}{2×3×4+2×3×4×2^{3}+2×3×4×4^{3}}$
=$\frac{4×5×6×(1+2^{3}+4^{3})}{2×3×4×(1+2^{3}+4^{3})}$
=5