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

A.1
B.3
C.6
D.7

1.C 【解析】因为7个有理数的积为负数,所以负因数的个数为奇数,故选C.

2[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.

2. 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.

3[2026 山东菏泽期中,中]观察下列各式:$\frac{1}{2}×\frac{2}{3}=\frac{1}{3}$,$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}=\frac{1}{4}$,$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×\frac{4}{5}=\frac{1}{5}$,….
(1) $(-\frac{1}{2})×(-\frac{2}{3})×(-\frac{3}{4})×…×(-\frac{9}{10})=$;
(2) 根据上面的规律计算:$(\frac{1}{100}-1)×(\frac{1}{99}-1)×(\frac{1}{98}-1)×…×(\frac{1}{2}-1)=$.

3. (1) $-\dfrac{1}{10}$ (2) $-\dfrac{1}{100}$ 【解析】(1) $(-\dfrac{1}{2})×(-\dfrac{2}{3})×(-\dfrac{3}{4})×…×(-\dfrac{9}{10}) = -\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{4}×…×\dfrac{9}{10} = -\dfrac{1}{10}$,故答案为$-\dfrac{1}{10}$.
(2) $(\dfrac{1}{100}-1)×(\dfrac{1}{99}-1)×(\dfrac{1}{98}-1)×…×(\dfrac{1}{2}-1) = (-\dfrac{99}{100})×(-\dfrac{98}{99})×(-\dfrac{97}{98})×…×(-\dfrac{1}{2}) = -\dfrac{99}{100}×\dfrac{98}{99}×\dfrac{97}{98}×…×\dfrac{1}{2} = -\dfrac{1}{100}$,故答案为$-\dfrac{1}{100}$.

4[2025河北保定期中,中]观察下图.

利用这种运算律可以得到$(6+4)×3=6×3+4×3=30$.
(1)它的计算过程可以解释这一运算律；
(2)请你利用这种运算律计算: $118 \frac{4}{5}×999+(-\frac{1}{5})×999-18 \frac{3}{5}×999$.
包会的鸭
刷素养→走向重高

4.【解】(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) $118 \dfrac{4}{5}×999+(-\dfrac{1}{5})×999-18 \dfrac{3}{5}×999 = (118 \dfrac{4}{5}-\dfrac{1}{5}-18 \dfrac{3}{5})×999 = (118 \dfrac{3}{5}-18 \dfrac{3}{5})×999 = 100×999 = 99\ 900.$

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) $(1+\frac{1}{2}+\frac{1}{3}+···+\frac{1}{n})×(\frac{1}{2}+\frac{1}{3}+···+\frac{1}{n+1})-(1+\frac{1}{2}+\frac{1}{3}+···+\frac{1}{n+1})×(\frac{1}{2}+\frac{1}{3}+···+\frac{1}{n}).$

5.【解】(1)设$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}$为$A$,
$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}$为$B$,则原式$=(1+A)·B-(1+B)A = B+AB-A-AB = B-A = \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}-(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}) = \dfrac{1}{7}.$
(2)设$\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{n}$为$A$,
$\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{n+1}$为$B$,则原式$=(1+A)B-(1+B)A = B+AB-A-AB = B-A = \dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{n+1}-(\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{n}) = \dfrac{1}{n+1}.$