Răspuns:
Explicație pas cu pas:
a) (3n+1)/20 -> fractie subunitara ireductibila
=> 3 n + 1 < 20, dar si 3 n + 1 ≠ 4, 10, 16
3 n < 20 - 1
n < 19/3
n < 6,(3)
n ∈ N => 0 ≤ n ≤ 6
daca n = 0 ⇒ ( 3n+1)/20 = (3×0+1)/20 = 1/20 → fractie subunitara ireductibila
n ≠ 1, deoarece ( 3×1+1)/20 = 4/20 nu este fractie ireductibila
n = 2 ⇒ (3×2+1)/20 = 7/20 → fractie ireductibila
n ≠ 3, deoarece (3×3+1)/20 = 10/20 nu este fractie ireductibila
n = 4 ⇒ ( 3×4+1)/20 = 13/20 → fractie subunitara ireductibila
n ≠ 5, deoarece (3×5+1)/20 = 16/20 nu este o fractie ireductibila
n = 6 ⇒ ( 3×6+1)/20 = 19/20 → fractie subunitara ireductibila
⇒ n = { 0, 2, 4, 6 } ∈ N
___________________________________________________
b)
18/(n+3) → fractie supraunitara si ireductibila
=> n + 3 < 18
n < 18 - 3
n < 15
n + 3 ≠ 3, 4, 6, 8, 9, 10, 12, 14, 15, 16
n ≠3-3; 4-3; 6-3; 8-3; 9-3; 10-3; 12-3; 14-3; 15-3; 16-3
=> n ≠ 0; 1; 3; 5; 6; 7; 9; 11; 12; 13
=> n = 2, 4, 8, 10 si 14 ∈ N
Verific:
18/(2+3) = 18/5 -> fractie supraunitara, ireductibila
18/(4+3) = 18/7
18/(8+3) = 18/11
18/(10+3) = 18/13
18/(14+3) = 18/17
[tex]\it a)\ \ \dfrac{3n+1}{20}<1 \Rightarrow 3n+1<20|_{-1} \Rightarrow 3n<19 \Rightarrow n\in\{0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6\}\\ \\ \\ \dfrac{3n+1}{20}\ ireductibil\breve{a} \Rightarrow 3n+1=impar \Rightarrow 3n=par \Rightarrow n=par\Rightarrow \\ \\ \Rightarrow n\in\{0,\ 2,\ 4,\ 6\}\\ \\ \\ b)\ \ \dfrac{18}{n+3}>1 \Rightarrow n+3<18|15_{-3} \Rightarrow n<15\Rightarrow n\in\{0,\ 1,\ 2,\ 3,\ ...,\ 14\}\\ \\ \dfrac{18}{n+3}\ ireductibil\breve{a} \Rightarrow n\in\{2,\ \ 4,\ \ 8,\ \ 10,\ \ 14\}[/tex]
