Question

Să se determine mulțimile:​

Answer 1

Explicație pas cu pas:

a)

$\dfrac{4n}{n + 2} = \dfrac{4(n + 2) - 8}{n + 2} = 4 - \dfrac{8}{n + 2} \in \mathbb{N}$

$(n + 2) \in \mathcal{D}_{8} \iff (n + 2) \in \Big\{ 1; 2; 4; 8\Big\} \ \ \Big|- 2 \\n \in \Big\{ - 1; 0; 2; 6\Big\} \cap \mathbb{N} \implies A = \Big\{0; 2; 6\Big\}$

b)

$\dfrac{6n + 7}{3n + 1} = \dfrac{2(3n + 1) + 5}{3n + 1} = 2 + \dfrac{5}{3n + 1} \in \mathbb{Z}$

$(3n + 1) \in \mathcal{D}_{5} \iff (3n + 1) \in \Big\{ -5; -1; 1; 5\Big\} \ \ \Big|- 1$

$3n \in \Big\{ - 6; -2; 0; 4\Big\}\ \ \Big|:3 \iff n \in \Big\{ - 2; - \dfrac{2}{3}; 0; \dfrac{4}{3}\Big\} \cap \mathbb{Z}$

$\implies B = \Big\{-2; 0\Big\}$

c)

$\dfrac{2n^{2} + 4n + 2}{n^{2} + 1} = \dfrac{2(n^{2} + 1) + 4n}{n^{2} + 1} = 2 + \dfrac{4n}{n^{2} + 1} \in \mathbb{N}$

$4n \geqslant n^{2} + 1 \iff n^{2} - 4n + 1 \leqslant 0 \\ n^{2} - 4n + 4 \leqslant 3 \iff {(n - 2)}^{2} \leqslant 3$

$n \in \mathbb{N} \implies n \in \Big\{0; 1; 2; 3\Big\}$

verificăm:

$\implies C = \Big\{0; 1\Big\}$