Question

Aflat elementele multimilor:A={x∈N/51 pe 4x-3∈N}
B={k∈N/3k+22 pe k+4∈N}

Answer 1

$A = \Big\{x\in\mathbb_{N} \big| \dfrac{51}{4x-3} \in\mathbb_{N} \Big\} \\ \\ 4x-3 \big | 51 \Rightarrow 4x-3\in D_{51}^+ \Rightarrow 4x - 3 \in\Big\{1;51\Big\} \Big|+3 \Rightarrow \\ \Rightarrow 4x \in\Big\{1+3; 51+3\Big\} \Rightarrow 4x\in \Big\{4;54\Big\}\Big|:4 \Rightarrow x \in\Big\{1; \dfrac{54}{4}\Big\} \\ \\ Dar, x\in\mathbb_{N} \Rightarrow \boxed{A = \Big\{1\Big\}}$


$B = \Big\{k\in\mathbb_{N} \Big| \dfrac{3k+22}{k+4}\in \mathbb_{N} \Big\} \\ \\ \dfrac{3k+22}{k+4} = \dfrac{3k+12+10}{k+4} = \dfrac{3\cdot(k+4)+10}{k+4} = \dfrac{3\cdot(k+4)}{k+4}+ \dfrac{10}{k+4} = \\ \\ =3 + \dfrac{10}{k+4} \in\mathbb_{N} \\ \\ \boxed{1}\quad\dfrac{10}{k+4} \geq -3 ; Dar k\in \mathbb_{N} \Rightarrow k+4\ \textgreater \ 0 \Rightarrow \dfrac{10}{k+4} \geq -3, \quad \forall k\in \mathbb_{N} \\ \\ \Rightarrow k\in \mathbb_{N}$

$\boxed{2} \quad k+4\in D_{10}^+ \Rightarrow k+4 \in \Big\{1;2;5;10\Big\}\Big|-4 \Rightarrow k\in\Big\{-3;-2;1;6\Big\} \\ Dar, k\in\mathbb_{N} \Rightarrow k\in\Big\{1,6\Big\} \\ \\ Din \boxed{1} \cap \boxed{2}, adica din: \mathbb_{N} \cap \Big\{1;6\Big\} \Rightarrow\boxed{ B = \Big\{1,6\Big\}}$