《2025年课堂点睛七年级数学上册人教版安徽专版》

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

答案：
(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：
(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：
(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=$\frac{1}{2}×\frac{3}{2}×\frac{2}{3}×\frac{4}{3}×\frac{3}{4}×\frac{5}{4}×…×\frac{2021}{2022}×\frac{2023}{2022}×\frac{2022}{2023}×\frac{2024}{2023}=\frac{1}{2}×\frac{2024}{2023}=\frac{1012}{2023}$.