Question

Am nevoie de ajutor
(n+3)!÷(n+2)!<5
Va rog urgent!!!!

Answer 1

(n+3)!÷(n+2)!<5

(n+3)!=(n+3) x (n+2)!

(n+3) x (n+2)!÷(n+2)!<5

n+3 <5

n<2

S=(-∞;2)

Answer 2

(n+3)! /(n+2)! <5

n+3≥0 si n+2≥0 => n ≥ - 2; n∈[-2, +∞)  

(n+3)(n+2)! /(n+2)! <5

n+3<5

n<2 => n∈(-∞, 2)  

(-∞, 2) ∩ [-2, +∞) = [-2, 2)

deci, n∈[-2, 2)