《2025年新坐标同步练习七年级数学上册人教版青海专用》

5. 计算：$99\dfrac{71}{72}× (-36)$.

答案： 解:原式=(100-$\frac{1}{72}$)×(-36)
=100×(-36)-$\frac{1}{72}$×(-36)
=-3 600+$\frac{1}{2}$=-3 599$\frac{1}{2}$.

6. 计算：$-2025\dfrac{1}{4}-2026\dfrac{2}{5}+4047\dfrac{2}{5}-1\dfrac{1}{2}$.

答案： 解:原式=
[-2 025+(-$\frac{1}{4}$)]+[-2 026+(-$\frac{2}{5}$)]+(4 047+$\frac{2}{5}$)+[-1+(-$\frac{1}{2}$)]
=[-2 025+(-2 026)+4 047+(-1)]+[(-$\frac{1}{4}$)+(-$\frac{2}{5}$)+$\frac{2}{5}$+(-$\frac{1}{2}$)]
=-5+(-$\frac{3}{4}$)=-5$\frac{3}{4}$.

7. 观察下面的变化规律：
$\dfrac{1}{1× 2}= 1-\dfrac{1}{2}$；$\dfrac{1}{2× 3}= \dfrac{1}{2}-\dfrac{1}{3}$；
$\dfrac{1}{3× 4}= \dfrac{1}{3}-\dfrac{1}{4}$；$\dfrac{1}{4× 5}= \dfrac{1}{4}-\dfrac{1}{5}$；
……
根据上面的规律,请你完成下列各题：
(1)$\dfrac{1}{2025× 2026}=$$\frac{1}{2025}-\frac{1}{2026}$,$\dfrac{1}{n(n+1)}=$$\frac{1}{n}-\frac{1}{n+1}$；
(2)计算：$\dfrac{1}{1× 2}+\dfrac{1}{2× 3}+\dfrac{1}{3× 4}+… +\dfrac{1}{9× 10}$；
$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{9×10}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{9}-\frac{1}{10}$
$=1-\frac{1}{10}=\frac{9}{10}$.
(3)计算：$\dfrac{1}{1× 3}+\dfrac{1}{3× 5}+\dfrac{1}{5× 7}+… +\dfrac{1}{2023× 2025}$.
$\frac{1}{1×3}+\frac{1}{3×5}+\frac{1}{5×7}+…+\frac{1}{2023×2025}$
$=\frac{1}{2}×(1-\frac{1}{3})+\frac{1}{2}×(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}×(\frac{1}{5}-\frac{1}{7})+…+\frac{1}{2}×(\frac{1}{2023}-\frac{1}{2025})$
$=\frac{1}{2}×[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+…+(\frac{1}{2023}-\frac{1}{2025})]$
$=\frac{1}{2}×(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+…+\frac{1}{2023}-\frac{1}{2025})$
$=\frac{1}{2}×(1-\frac{1}{2025})$
$=\frac{1}{2}×\frac{2024}{2025}=\frac{1012}{2025}$.

答案： 解:
(1)$\frac{1}{2025}$-$\frac{1}{2026}$ $\frac{1}{n}$-$\frac{1}{n+1}$
(2)$\frac{1}{1×2}$+$\frac{1}{2×3}$+$\frac{1}{3×4}$+…+$\frac{1}{9×10}$
=1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+…+$\frac{1}{9}$-$\frac{1}{10}$
=1-$\frac{1}{10}$=$\frac{9}{10}$.
(3)$\frac{1}{1×3}$+$\frac{1}{3×5}$+$\frac{1}{5×7}$+…+$\frac{1}{2023×2025}$
=$\frac{1}{2}$×(1-$\frac{1}{3}$)+$\frac{1}{2}$×($\frac{1}{3}$-$\frac{1}{5}$)+$\frac{1}{2}$×($\frac{1}{5}$-$\frac{1}{7}$)+…+$\frac{1}{2}$×($\frac{1}{2023}$-$\frac{1}{2025}$)
=$\frac{1}{2}$×[(1-$\frac{1}{3}$)+($\frac{1}{3}$-$\frac{1}{5}$)+($\frac{1}{5}$-$\frac{1}{7}$)+…+($\frac{1}{2023}$-$\frac{1}{2025}$)]
=$\frac{1}{2}$×(1-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+$\frac{1}{5}$-$\frac{1}{7}$+…+$\frac{1}{2023}$-$\frac{1}{2025}$)
=$\frac{1}{2}$×(1-$\frac{1}{2025}$)
=$\frac{1}{2}$×$\frac{2024}{2025}$=$\frac{1012}{2025}$.