《2025年课堂点睛七年级数学上册人教版宁夏专版》

5. 计算：
(1)$-2\frac {3}{4}+(-3\frac {2}{3})+20\frac {1}{4}-7\frac {1}{6}$；
(2)$(-15\frac {5}{6})+(-19\frac {3}{4})+14\frac {2}{3}+(-1\frac {1}{2})$.

答案： 5.
(1)解:原式$=(-2\frac{3}{4})+(-3\frac{2}{3})+20\frac{1}{4}+(-7\frac{1}{6})=(-2)+(-\frac{3}{4})+(-3)+(-\frac{2}{3})+20+\frac{1}{4}+(-7)+(-\frac{1}{6})=[(-2)+(-3)+20+(-7)]+[(-\frac{3}{4})+(-\frac{2}{3})+\frac{1}{4}+(-\frac{1}{6})]=8+(-\frac{4}{3})=6\frac{2}{3}. $
(2)解:原式$=(-15-\frac{5}{6})+(-19-\frac{3}{4})+(14+\frac{2}{3})+(-1-\frac{1}{2})=(-15-19+14-1)+(-\frac{5}{6}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2})=-21-\frac{17}{12}=-22\frac{5}{12}.$

6. 计算：
(1)$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+... +\frac {1}{2024×2025}$；
(2)$\frac {1}{2}-\frac {1}{6}-\frac {1}{12}-\frac {1}{20}-\frac {1}{30}$.

答案： 6.
(1)解:原式$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots+\frac{1}{2024}-\frac{1}{2025}=1-\frac{1}{2025}=\frac{2024}{2025}. $
(2)解:原式$=\frac{1}{2}-\frac{1}{2×3}-\frac{1}{3×4}-\frac{1}{4×5}-\frac{1}{5×6}=\frac{1}{2}-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{3}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{5})-(\frac{1}{5}-\frac{1}{6})=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}=\frac{1}{6}.$