《2025年节节高大象出版社七年级数学上册华师大版》

23. (11分)定义新运算：$a*b=\dfrac{1}{a}-\dfrac{1}{b}$，$a\otimes b=\dfrac{1}{ab}$.
例如，$3*7=\dfrac{1}{3}-\dfrac{1}{7}=\dfrac{4}{21}$，$3\otimes 7=\dfrac{1}{3× 7}=\dfrac{1}{21}$.
若$a*b = a\otimes b$，则称有理数$a$，$b$为一组“魅力数对”.
例如，$2*3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$，$2\otimes 3=\dfrac{1}{2× 3}=\dfrac{1}{6}$，$2*3 = 2\otimes 3$，所以2，3就是一组“魅力数对”.
(1)下列各组数是“魅力数对”的是①③(填序号).
①$a = 1$，$b = 2$；②$a = -1$，$b = 2$；③$a = -\dfrac{4}{3}$，$b = -\dfrac{1}{3}$.
(2)请再写出一组“魅力数对”：$3$，$4$（答案不唯一）.
(3)计算：$2025*2026 - 2025\otimes 2026$.

答案： 23.
(1)①③ [解析]因为$a*b=\frac{1}{a}-\frac{1}{a× b}$，$a\otimes b=\frac{1}{a× b}$，所以$1*2 = \frac{1}{1} - \frac{1}{1×2} = \frac{1}{2}$，$1\otimes2 = \frac{1}{1×2} = \frac{1}{2}$，
所以$1*2 = 1\otimes2$.
所以$a = 1$，$b = 2$是“魅力数对”.
因为$(-1)*2 = \frac{1}{-1} - \frac{1}{(-1)×2} = - \frac{3}{2}$，$(-1)\otimes2 = \frac{1}{(-1)×2} = - \frac{1}{2}$，
所以$(-1)*2 \neq (-1)\otimes2$.
所以$a = - 1$，$b = 2$不是“魅力数对”.
因为$(-\frac{4}{3})*(-\frac{1}{3}) = \frac{1}{-\frac{4}{3}} - \frac{1}{(-\frac{4}{3})×(-\frac{1}{3})} = \frac{9}{4}$，$(-\frac{4}{3})\otimes(-\frac{1}{3}) = \frac{1}{(-\frac{4}{3})×(-\frac{1}{3})} = \frac{9}{4}$，
所以$(-\frac{4}{3})*(-\frac{1}{3}) = (-\frac{4}{3})\otimes(-\frac{1}{3})$.
所以$a = -\frac{4}{3}$，$b = -\frac{1}{3}$是“魅力数对”.
故答案为①③.
(2)$3$，$4$（答案不唯一） [解析]因为$3*4 = \frac{1}{3} - \frac{1}{3×4} = \frac{1}{12}$，$3\otimes4 = \frac{1}{3×4} = \frac{1}{12}$，
所以$3*4 = 3\otimes4$.
所以$3$，$4$是“魅力数对”.
故答案为$3$，$4$（答案不唯一）.
(3)$2025*2026 - 2025\otimes2026$
$= \frac{1}{2025} - \frac{1}{2025×2026} - \frac{1}{2025×2026}$
$= \frac{2026 - 2025}{2025×2026} - \frac{1}{2025×2026}$
$= 0$.