[tex]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\}}[/tex]
[tex]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}[/tex]
[tex]\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\}}[/tex]
