《2025年课堂点睛七年级数学上册人教版宁夏专版》

4. 计算：
(1)$-105\frac {5}{9}×(-\frac {3}{5})$；
(2)$91\frac {15}{16}×(-8)$。

答案： 4.
(1)解:原式$=105\frac {5}{9}×\frac {3}{5}=105×\frac {3}{5}+\frac {5}{9}×\frac {3}{5}=63\frac {1}{3}$.
(2)解:原式$=(92-\frac {1}{16})×(-8)=92×(-8)-\frac {1}{16}×(-8)=-736+\frac {1}{2}=-735\frac {1}{2}$.

5. 计算：
(1)$\frac {1}{12}÷(-\frac {1}{6}-\frac {2}{3}+\frac {1}{4})$；
(2)$(-\frac {1}{63})÷(\frac {1}{9}-\frac {2}{7}+\frac {2}{3})$。

答案： 5.
(1)解:
∵$(-\frac {1}{6}-\frac {2}{3}+\frac {1}{4})÷\frac {1}{12}=(-\frac {1}{6}-\frac {2}{3}+\frac {1}{4})×12=-\frac {1}{6}×12-\frac {2}{3}×12+\frac {1}{4}×12=-2-8+3=-7$,
∴原式$=-\frac {1}{7}$.
(2)解:
∵$(\frac {1}{9}-\frac {2}{7}+\frac {2}{3})÷(-\frac {1}{63})=(\frac {1}{9}-\frac {2}{7}+\frac {2}{3})×(-63)=(-7)+18-42=-31$,
∴原式$=-\frac {1}{31}$.

6. 计算：
(1)$1-3×2+5+7-9×2+11+13-15×2+17+…[phone]×2+2027$；
(2)$(\frac {1}{2}×\frac {3}{2})×(\frac {2}{3}×\frac {4}{3})×(\frac {3}{4}×\frac {5}{4})×…×(\frac {2021}{2022}×\frac {2023}{2022})×(\frac {2022}{2023}×\frac {2024}{2023})$。

答案： 6.
(1)解:原式=(1-3×2+5)+(7-9×2+11)+(13-15×2+17)+…+(2023-2025×2+2027)=0+0+0+…+0=0.
(2)解:原式$=\frac {1}{2}×\frac {3}{2}×\frac {2}{3}×\frac {4}{3}×\frac {3}{4}×\frac {5}{4}×…×\frac {2021}{2022}×\frac {2023}{2022}×\frac {2022}{2023}×\frac {2024}{2023}=\frac {1}{2}×\frac {2024}{2023}=\frac {1012}{2023}$.