《2025年亮点给力大试卷七年级数学上册苏科版》

23. (4分)若A,B,C三点在数轴上表示的数分别为a,b,c,且满足$|a-b|= 2|a-c|$,则称B为A,C两点的倍距点.例如:若$a= -2,b= 4,c= 1$,因为$|a-b|= |-2-4|= 6,2|a-c|= 2×|-2-1|= 6$,所以$|a-b|= 2|a-c|$,即B是A,C两点的倍距点.
(1) 若$a= 3,b= -5,c= -1$,请说明:B是A,C两点的倍距点;
(2) 若$a= -1$,B是A,C两点的倍距点,且$a-c= 2$,求b的值.

答案：
(1)因为a = 3,b = - 5,c = - 1,所以|a - b| = |3 - (-5)| = 8,2|a - c| = 2×|3 - (-1)| = 2×4 = 8,即|a - b| = 2|a - c|.所以B是A,C两点的倍距点;
(2)因为B是A,C两点的倍距点,所以|a - b| = 2|a - c|.因为a - c = 2,a = - 1,所以|-1 - b| = 4.所以 - 1 - b = 4或 - 1 - b = - 4,解得b = - 5或b = 3.所以b = - 5或3.

24. (4分)新素养 运算能力 (2025·江苏扬州期末)阅读:$\frac {1}{2}×\frac {2}{3}= \frac {1}{3},\frac {1}{2}×\frac {2}{3}×\frac {3}{4}= \frac {1}{4},\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}= \frac {1}{5},...,\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×... ×\frac {n}{n+1}= \frac {1}{n+1}$(n为正整数).
试根据以上规律,解答下列问题:
(1) 计算:$(\frac {1}{2}-1)×(\frac {1}{3}-1)×(\frac {1}{4}-1)×... ×(\frac {1}{100}-1)$;
(2) 将2025减去它的$\frac {1}{2}$,再减去余下的$\frac {1}{3}$,再减去余下的$\frac {1}{4}$,再减去余下的$\frac {1}{5}$……以此类推,直到减去余下的$\frac {1}{2025}$,最后的结果是多少?

答案：
(1)原式 = (-$\frac{1}{2}$)×(-$\frac{2}{3}$)×(-$\frac{3}{4}$)×…×(-$\frac{99}{100}$)= - ($\frac{1}{2}$×$\frac{2}{3}$×$\frac{3}{4}$×…×$\frac{99}{100}$)= - $\frac{1}{100}$;
(2)由题意,得2025×(1 - $\frac{1}{2}$)×(1 - $\frac{1}{3}$)×(1 - $\frac{1}{4}$)×…×(1 - $\frac{1}{2025}$)=2025×($\frac{1}{2}$×$\frac{2}{3}$×$\frac{3}{4}$×…×$\frac{2024}{2025}$)=2025×$\frac{1}{2025}$ = 1.则最后的结果是1.