Answer:
[tex] \red{\bf A= \{2,4 \}}[/tex]
Step-by-step explanation:
[tex] \bf A = \bigg \{x \in \mathbb{N} \bigg| \: \dfrac{x + 2}{x - 1} \in \mathbb{N} \bigg \}[/tex]
[tex]\bf x - 1 \: \: \: \bigg| \: \: \: x + 2[/tex]
[tex]\bf \underline{x - 1 \: \bigg| \: x - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }( - )[/tex]
[tex]\bf x - 1 \: \bigg| \: x + 2 - (x - 1)[/tex]
[tex]\bf x - 1 \: \bigg| \: x + 2 - x + 1[/tex]
[tex]\bf x - 1 \: \big| \:3 \implies x - 1 \in D_{3}[/tex]
[tex]\bf x - 1 \in \{1,3 \} \: \: \: \: \bigg| + 1[/tex]
[tex]\bf x \in \{2,4 \} \implies \red{\boxed{ \bf A= \{2,4 \}}}[/tex]
