pr KASZAA: oblicz

n 3		n+2 4
	:		= 1:5

dochodzę do postaci n² − 2n + 12=0 i delta wychodzi ujemna

sushi_gg6397228: zapisz obliczenia

Lukas:

n!		(n+2)!
	:
3!(n−3)!		4!(n−2)!

n!		n!(n+1)(n+2)
	:
n!(n−1)(n−2)(n−3)*3!		n!(n−1)(n−2)*4!

1		(n+1)(n+2)
	:
(n−1)(n−2)(n−3)*3!		(n−1)(n−2)*4!

4!
(n+1)(n+2)(n−3)

Lewą stronę tylko robiłem

razor:

n 3		1
	=
n+2 4		5

	n 3		n+2 4
5		=

	n!		(n+2)!
5*		=
	3!*(n−3)!		4!(n−2)!

	n(n−1)(n−2)		(n−1)n(n+1)(n+2)
5*		=
	6		24

20n(n−1)(n−2) = (n−1)n(n+1)(n+2) 20n(n−1)(n−2) − (n−1)n(n+1)(n+2) = 0 n(n−1)[20(n−2)−(n+1)(n+2)] = 0 n(n−1)[20n−40−n²−3n−2] = 0 n = 0 lub n = 1 lub n²−17n+42 = 0 ⇔ (n−14)(n−3) = 0 ⇔ n = 14 lub n = 3 połącz to z dziedziną i masz wynik

KASZAA:

(n−3)! (n−2) (n−1) n		(n+2)!		1
	:		=
6(n−3)!		4!(n+2)!		5

(n−2) (n−1) n		(n−2)! (n−1) (n+1) n (n+2)		1
	:		=
6		24(n+2)!		5

(n−2) (n−1) n		(n−1) (n+1) n (n+2)		1
	*		=
6		24		5

n−2		1
	=
(n+1) (n+2)		5

5(n−2) = (n+1)(n+2) n2 − 2n + 12=0

Janek191:

n 3		n !		(n −2)*( n −1)*n
	=		=
		3 ! *( n − 3) !		6

n + 2 4		( n + 2) !		( n −1)*n*( n + 1)*(n + 2)
	=		=
		4 ! *( n − 2) !		24

pigor: ..., lub... tak :

n 3		n+2 4
	:		= 1 : 5 i n ≥3 ⇒

	n(n−1)(n−2)		4*3*2*1		1
⇒		*		=		⇔
	3*2*1		(n+2)(n+1)n(n−1)		5

	4(n−2)		1
⇔		=		⇔ n2+3n+2= 20n−40 ⇔
	(n+2)(n+1)		5

⇔ n²−17n+42= 0 ⇔ n²−3n−14n+42= 0 ⇔ n(n−3)−14(n−3)= 0 ⇔ ⇔ (n−3)(n−14)= 0 ⇔ n=3 v n=14 ⇔ n∊{3,14} . ...