《2025年课堂作业武汉出版社七年级数学上册人教版》

B级
8.计算:$2023×20252025-2025×20242024=$-20252025.

答案： -20 252 025

9.计算.
(1)$49\frac{24}{25}×(-5)$.
(2)$325×(-125)×(-11)×8×\frac{1}{25}×\left(-\frac{5}{11}\right)$.
(3)$-7.8×(-8.1)×0×|-19.6|$.
(4)$-|-0.25|×(-5)×4×\left(-\frac{1}{25}\right)$.

答案： 9.
(1)原式$=\left(50-\frac{1}{25}\right)×(-5)=-50×5+\frac{1}{25}×5=-249\frac{4}{5}$.
(2)原式$=(-11)×\left(-\frac{5}{11}\right)×\frac{1}{25}×(-125)×8×325=-65000$.
(3)原式=0.
(4)原式$=-0.25×(-5)×4×\left(-\frac{1}{25}\right)=-\frac{1}{5}$.

10.将2025减去它的$\frac{1}{2}$,再减去余下的$\frac{1}{3}$,再减去余下的$\frac{1}{4}$……依次类推,直至减去余下的$\frac{1}{2025}$,最后的得数是多少?

答案： 1

11.观察下面按规律排成的一列数(已写出了第1至16个数):$\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\frac{1}{4},\frac{2}{3},\frac{3}{2},\frac{4}{1},\frac{1}{5},\frac{2}{4},\frac{3}{3},\frac{4}{2},\frac{5}{1},\frac{1}{6},…$.
(1)依此规律,写出第17,18,19个数,分别为______.
(2)若某一个数为$\frac{2}{a}(a\geq3且a$为整数),则$\frac{2}{a}$的前一个数为______;$\frac{2}{a}$的后一个数为______.
(3)在上面这列数中,从左起第$m个数记为F(m)$,当$F(m)= \frac{2}{9999}$时,求$m的值和这m$个数的积.
(1)依此规律,写出第17,18,19个数,分别为$\frac{2}{5}$,$\frac{3}{4}$,$\frac{4}{3}$.
(2)若某一个数为$\frac{2}{a}(a\geq3且a$为整数),则$\frac{2}{a}$的前一个数为$\frac{1}{a+1}$;$\frac{2}{a}$的后一个数为$\frac{3}{a-1}$.
(3)在上面这列数中,从左起第$m个数记为F(m)$,当$F(m)= \frac{2}{9999}$时,求$m的值和这m$个数的积.
$m=1+2+3+\cdots+999+2=\frac{999×(999+1)}{2}+2=499502$,这m个数的积$=\frac{1}{1}×\left(\frac{1}{2}×\frac{2}{1}\right)×\left(\frac{1}{3}×\frac{2}{2}×\frac{3}{1}\right)×\left(\frac{1}{4}×\frac{2}{3}×\frac{3}{2}×\frac{4}{1}\right)×\left(\frac{1}{5}×\frac{2}{4}×\frac{3}{3}×\frac{4}{2}×\frac{5}{1}\right)×\cdots×\frac{1}{1000}×\frac{2}{999}=\frac{1}{499500}$.

答案： 11.
(1)$\frac{2}{5}$,$\frac{3}{4}$,$\frac{4}{3}$
(2)$\frac{1}{a+1}$ $\frac{3}{a-1}$
(3)$m=1+2+3+\cdots+999+2=\frac{999×(999+1)}{2}+2=499502$,这m个数的积$=\frac{1}{1}×\left(\frac{1}{2}×\frac{2}{1}\right)×\left(\frac{1}{3}×\frac{2}{2}×\frac{3}{1}\right)×\left(\frac{1}{4}×\frac{2}{3}×\frac{3}{2}×\frac{4}{1}\right)×\left(\frac{1}{5}×\frac{2}{4}×\frac{3}{3}×\frac{4}{2}×\frac{5}{1}\right)×\cdots×\frac{1}{1000}×\frac{2}{999}=\frac{1}{499500}$.