《2025年初中必刷题七年级数学上册人教版》

1[2025湖北武汉质检,中]七个有理数的积为负数,其中负因数的个数一定不可能是(C)
A.1
B.3
C.6
D.7

答案： C 【解析】因为七个有理数的积为负数,所以负因数的个数为奇数,故选C.

如图,阶梯图的每个台阶上都标着一个数,从下到上的第1个至第3个台阶上依次标着$-2,-1,-\frac {1}{2}$,且任意相邻三个台阶上数的积都相等.对下列结论判断正确的是( B )
结论Ⅰ:从下到上前2022个台阶上的数的积是-1;
结论Ⅱ:数$-\frac {1}{2}$所在的台阶数用正整数k表示为3k.
A.Ⅰ对、Ⅱ错
B.Ⅰ错、Ⅱ对
C.Ⅰ和Ⅱ都对
D.Ⅰ和Ⅱ都错

答案： B 【解析】因为任意相邻三个台阶上数的积都相等,且$-2×(-1)×(-\frac {1}{2})=-1,2022=674×3,$所以从下到上前2022个台阶上的数的积是1,故结论Ⅰ错误;由题意得,数$-\frac {1}{2}$在第3个台阶上,第6个台阶上,第9个台阶上,…,所以数$-\frac {1}{2}$所在的台阶数用正整数k表示为3k,故结论Ⅱ正确.故选B.

3[2024湖北荆州期末,中]如果4个不等的偶数m,n,p,q满足$(3-m)(3-n)(3-p)(3-q)= 9$,那么$m+n+p+q$等于12.

答案： 12 【解析】因为m,n,p,q是4个不等的偶数,所以(3-m),(3-n),(3-p),(3-q)均为不等的奇数.因为9=3×1×(-1)×(-3),所以可令3-m=3,3-n=1,3-p=-1,3-q=-3,解得m=0,n=2,p=4,q=6,所以m+n+p+q=0+2+4+6=12.故答案为12.

(1)它的计算过程可以解释分配律这一运算律;
(2)请你利用这种运算律计算:$118\frac {4}{5}×999+(-\frac {1}{5})×999-18\frac {3}{5}×999.$
$118\frac {4}{5}×999+(-\frac {1}{5})×999-18\frac {3}{5}×999=(118\frac {4}{5}-\frac {1}{5}-18\frac {3}{5})×999=(118\frac {3}{5}-18\frac {3}{5})×999=100×999=99900$

答案：
(1)它的计算过程可以解释分配律这一运算律,故答案为分配律$.(2)118\frac {4}{5}×999+(-\frac {1}{5})×999-18\frac {3}{5}×999=(118\frac {4}{5}-\frac {1}{5}-18\frac {3}{5})×999=(118\frac {3}{5}-18\frac {3}{5})×999=100×999=99900.$

5思想方法整体思想[较难]计算:$(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})-(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}).$
小明同学的解法如下:
解:设$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4})$为A,$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})$为B,则原式$=B(1+A)-A(1+B)= B+AB-A-AB= B-A= \frac {1}{5}$.请用上面方法计算:
(1)$(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7})-(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6});$
(2)$(1+\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n})(\frac {1}{2}+\frac {1}{3}+... +\frac {1}{n+1})-(1+\frac {1}{2}+\frac {1}{3}... +\frac {1}{n+1})(\frac {1}{2}+\frac {1}{3}... +\frac {1}{n}).$

答案：
(1)设$(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})$为$A,(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7})$为B.原式$=(1+A)B-(1+B)A=B+AB-A-AB=B-A=\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}-(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})=\frac {1}{7}.(2)$设$(\frac {1}{2}+\frac {1}{3}+\cdots +\frac {1}{n})$为$A,(\frac {1}{2}+\frac {1}{3}+\cdots +\frac {1}{n+1})$为B.原式$=(1+A)B-(1+B)A=B+AB-A-AB=B-A=(\frac {1}{2}+\frac {1}{3}+\cdots +\frac {1}{n+1})-(\frac {1}{2}+\frac {1}{3}+\cdots +\frac {1}{n})=\frac {1}{n+1}.$