《2025年拔尖特训七年级数学上册苏科版》

4. 计算：
(1)$2\frac{1}{3}-\left(+10\frac{1}{3}\right)+\left(-8\frac{1}{5}\right)-3\frac{2}{5}$。
(2)$\left(-17\frac{2}{3}\right)+16\frac{3}{4}+\left(-15\frac{1}{3}\right)-2\frac{1}{2}$。
(3)$-8721+53\frac{19}{21}-1279-43\frac{19}{21}$。
(4)$\frac{1}{-1×3}+\frac{1}{-3×5}+\frac{1}{-5×7}+\frac{1}{-7×9}+…+\frac{1}{-2021×2023}$。
(5)$\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}$。

答案： （1）原式=$2+\frac{1}{3}-10-\frac{1}{3}-8-\frac{1}{5}-3-\frac{2}{5}=(2-10-8-3)+(\frac{1}{3}-\frac{1}{3}-\frac{1}{5}-\frac{2}{5})=-19-\frac{3}{5}=-19\frac{3}{5}$.
（2）原式=[(-17)+16+(-15)+(-2)]+$[(-\frac{2}{3})+\frac{3}{4}+(-\frac{1}{3})+(-\frac{1}{2})]=-18+(-\frac{3}{4})=-18\frac{3}{4}$.
（3）原式=$-8721+53+\frac{19}{21}-1279-43-\frac{19}{21}=(-8721+53-1279-43)+(\frac{19}{21}-\frac{19}{21})=-9990$.
（4）原式=$-\frac{1}{1×3}-\frac{1}{3×5}-\frac{1}{5×7}-\frac{1}{7×9}-…-\frac{1}{2021×2023}=-\frac{1}{2}×(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+…+\frac{1}{2021}-\frac{1}{2023})=-\frac{1}{2}×(1-\frac{1}{2023})=-\frac{1}{2}×\frac{2022}{2023}=-\frac{1011}{2023}$.
（5）原式=$(1-\frac{1}{12})+(1-\frac{1}{20})+(1-\frac{1}{30})+(1-\frac{1}{42})+(1-\frac{1}{56})+(1-\frac{1}{72})=(1-\frac{1}{3×4})+(1-\frac{1}{4×5})+(1-\frac{1}{5×6})+(1-\frac{1}{6×7})+(1-\frac{1}{7×8})+(1-\frac{1}{8×9})=6-(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9})=6-(\frac{1}{3}-\frac{1}{9})=6-\frac{2}{9}=5\frac{7}{9}$.

5. 计算：
(1)$\frac{1}{30}÷\left(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}-\frac{2}{5}\right)$。
(2)$\left(-\frac{1}{56}\right)÷\left(\frac{1}{7}+\frac{3}{28}-\frac{1}{4}+\frac{5}{8}\right)$。

答案： （1）原式的倒数=$(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}-\frac{2}{5})÷\frac{1}{30}=(\frac{2}{3}-\frac{1}{10}+\frac{1}{6}-\frac{2}{5})×30=\frac{2}{3}×30-\frac{1}{10}×30+\frac{1}{6}×30-\frac{2}{5}×30=20-3+5-12=10$.所以原式=$\frac{1}{10}$.
（2）原式的倒数=$(\frac{1}{7}+\frac{3}{28}-\frac{1}{4}+\frac{5}{8})÷(-\frac{1}{56})=(\frac{1}{7}+\frac{3}{28}-\frac{1}{4}+\frac{5}{8})×(-56)=\frac{1}{7}×(-56)+\frac{3}{28}×(-56)-\frac{1}{4}×(-56)+\frac{5}{8}×(-56)=-8-6+14-35=-35$.所以原式=$-\frac{1}{35}$.

6. 计算：
(1)$\frac{2}{5}-\left|-1\frac{1}{2}\right|-\left(+2\frac{1}{4}\right)-$
$(-2.75)$。
(2)$-32\frac{1}{3}-\left[5\frac{1}{4}-\left(+3\frac{1}{7}\right)-3.25-2\frac{6}{7}\right]$。
(3)$\left|\frac{1}{3}-1\right|+\left|\frac{1}{5}-\frac{1}{3}\right|+\left|\frac{1}{7}-\frac{1}{5}\right|+…+\left|\frac{1}{97}-\frac{1}{95}\right|+\left|\frac{1}{99}-\frac{1}{97}\right|$。

答案： （1）原式=$\frac{2}{5}-1\frac{1}{2}-2\frac{1}{4}+2.75=\frac{2}{5}-1-\frac{1}{2}+(2.75-2\frac{1}{4})=\frac{2}{5}-1-\frac{1}{2}+\frac{1}{2}=\frac{2}{5}-1+(-\frac{1}{2}+\frac{1}{2})=\frac{2}{5}-1=-\frac{3}{5}$.
（2）原式=$-32\frac{1}{3}-[(5\frac{1}{4}-3.25)+(-3\frac{1}{7}-2\frac{6}{7})]=-32\frac{1}{3}-[2+(-6)]=-32\frac{1}{3}-(-4)=-32\frac{1}{3}+4=-28\frac{1}{3}$.
（3）原式=$1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+…+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}$.