a) A = \(\dfrac{6n+7}{2n+3}\) = \(\dfrac{6n+9}{2n+3}\) − \(\dfrac{2}{2n+3}\) nguyên
⇔ 2n + 3 ∈ Ư(2) = {-2; -1; 1; 2}
⇔ 2n ∈ {-5; -4; -2; -1}
Vì n nguyên nên n ∈ {-2; -1}
Bài 2:
a) Để B nguyên thì \(6n+7⋮2n+3\)
\(\Leftrightarrow-2⋮2n+3\)
\(\Leftrightarrow2n+3\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow2n\in\left\{-2;-4\right\}\)
hay \(n\in\left\{-1;-2\right\}\)