Question

Exercițiul 5 va rog!! ​

Answer 1

Răspuns:

a)

$x = \frac{n+10}{n+1} =\frac{(n+1)+9}{n+1} =1+\frac{9}{n+1}$

cum x ∈ Z ⇒  $\frac{9}{n+1}$ ∈ Z

adica (n+1) | 9

(n+1) ∈ {-9, -3, -1, 1, 3, 9}

n ∈ {-10, -4, -2, 0, 2, 8}

b)

$x = \frac{3n+14}{n+2} =\frac{3(n+2)+8}{n+2} =3+\frac{8}{n+2}$

cum x ∈ Z ⇒  $\frac{8}{n+2}$ ∈ 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)

$x = \frac{4n+7}{2n-1} =\frac{2(2n-1)+9}{2n-1} =2+\frac{9}{2n-1}$

cum x ∈ Z ⇒  $\frac{9}{2n-1}$ ∈ 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}

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