Răspuns:
a)
[tex]x = \frac{n+10}{n+1} =\frac{(n+1)+9}{n+1} =1+\frac{9}{n+1}[/tex]
cum x ∈ Z ⇒ [tex]\frac{9}{n+1}[/tex] ∈ Z
adica (n+1) | 9
(n+1) ∈ {-9, -3, -1, 1, 3, 9}
n ∈ {-10, -4, -2, 0, 2, 8}
b)
[tex]x = \frac{3n+14}{n+2} =\frac{3(n+2)+8}{n+2} =3+\frac{8}{n+2}[/tex]
cum x ∈ Z ⇒ [tex]\frac{8}{n+2}[/tex] ∈ Z
adica (n+2) | 8
(n+2) ∈ {-8, -4, -2, -1, 1, 2, 4, 8}
n ∈ {-10, -6, -4, -3, -1, 0, 2, 6}
c)
[tex]x = \frac{4n+7}{2n-1} =\frac{2(2n-1)+9}{2n-1} =2+\frac{9}{2n-1}[/tex]
cum x ∈ Z ⇒ [tex]\frac{9}{2n-1}[/tex] ∈ Z
adica (2n-1) | 9
(2n-1) ∈ {-9, -3, -1, 1, 3, 9}
2n ∈ {-8, -2, 0, 2, 4, 10}
n ∈ {-4, -1, 0, 1, 2, 5}
d)
vom folosi proprietatea: daca x | a si x | b atunci x | (a-b)
(2n + 1) | (3n + 8)
(2n + 1) | (2n + 1)
(2n + 1) | (n + 7)
(2n + 1) | (2n + 1)
(2n + 1) | (n - 6)
(2n + 1) | (n + 7)
(2n + 1) | 13
(2n + 1) ∈ {-13, -1, 1, 13}
2n ∈ {-14, -2, 0, 12}
n ∈ {-7, -1, 0, 6}
Explicație pas cu pas:
