
N!/(n-4)! = 22n!/(n-3)!
Ответы на вопрос

n(n-1)(n-2)(n-3)(n-4)! 22n(n-1)(n-2)(n-3)!
----------------------------- = ---------------------------
(n-4)! (n-3)!
n-3=22
n=25
Ответ: 25

To solve the equation , we can start by simplifying both sides of the equation.
First, let's simplify the factorials:
(We didn't expand the last term to keep it in a factorial form)
Now, rewrite the equation with these simplifications:
Next, we can cancel out the common terms in the numerator and denominator on both sides:
Now, let's consider two cases:
Case 1: (This means )
If , we can cancel out from both sides of the equation:
Now, we have:
Now, we can divide both sides by (assuming ):
Now, we have a simple equation for :
However, must be a positive integer, but is not an integer. Therefore, there are no solutions in positive integers for this case.
Case 2:
If , then the equation becomes:
Simplifying further:
Since and , both sides of the equation are equal to 6. So, the solution in this case is .
Therefore, the only solution in positive integers is .
Похожие вопросы







Топ вопросов за вчера в категории Алгебра









Последние заданные вопросы в категории Алгебра






-
Математика
-
Литература
-
Алгебра
-
Русский язык
-
Геометрия
-
Английский язык
-
Химия
-
Физика
-
Биология
-
Другие предметы
-
История
-
Обществознание
-
Окружающий мир
-
География
-
Українська мова
-
Информатика
-
Українська література
-
Қазақ тiлi
-
Экономика
-
Музыка
-
Право
-
Беларуская мова
-
Французский язык
-
Немецкий язык
-
МХК
-
ОБЖ
-
Психология
-
Физкультура и спорт
-
Астрономия
-
Кыргыз тили
-
Оʻzbek tili