to są zadania z silnią!!
rozwiąż nierównośc :
(n+2)! > 12n!
(n+2)! > 12n! , n ∈ N
n! · (n +1)(n +2) > 12n!
(n + 1)(n +2) > 12
n² + 2n + n + 2 - 12 > 0
n² +3n - 10 > 0
Δ = 9 - 4 · 1 · (-10) = 9 + 40 = 49, √Δ = 7
n₁ =
n₂ =
n ∈ (-∞ , -5) (2, +∞)
n ∈ N
Odp. n ∈ (2 , +∞)
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(n+2)! > 12n! , n ∈ N
n! · (n +1)(n +2) > 12n!
(n + 1)(n +2) > 12
n² + 2n + n + 2 - 12 > 0
n² +3n - 10 > 0
Δ = 9 - 4 · 1 · (-10) = 9 + 40 = 49, √Δ = 7
n₁ =
n₂ =
n ∈ (-∞ , -5) (2, +∞)
n ∈ N
Odp. n ∈ (2 , +∞)