Răspuns:
Explicație pas cu pas:
a)
[tex]\frac{4n+7}{2n-1} =\frac{2(2n-1)+9}{2n-1} =\frac{2(2n-1)}{2n-1}+\frac{9}{2n-1}=\\ \\ \\=2+\frac{9}{2n-1}[/tex]
2n - 1 ∈ {-9, -3, -1, 1, 3, 9} ⇔ 2n ∈ {-8, -2, 0, 2, 4, 10}
⇔ n ∈ {-4, -1, 0, 1, 2, 5}, n ≠ 0
⇒ n ∈ {-4, -1, 1, 2, 5}
b)
[tex]\frac{7n+13}{2n-1} =\frac{3(2n-1) +n+16}{2n-1} =3+\frac{n+16}{2n-1}[/tex]
n + 16 ≥ 2n - 1 ⇔ n ≤ 17 & (n + 16) | (2n - 1)
n = 17: 33 | 33
n = 6: 22 | 11
n = 2: 18 | 3
n = 1: 17 | 1
n = 0: 16 | -1, nevalid n ≠ 0
n = -1: 15 | -3
n = -5: 11 | -11
⇒ n ∈ {-5, -1, 1, 2, 6, 17}
