$a) Đặt$ $A = \frac{5n -3}{n +2} = 5 - \frac{13}{n +2}$
$Để$ $A ∈ Z ⇔ \frac{-13}{n +2} ∈ Z$
$Mà$ $n ∈ Z$
$⇒ n +2 ∈ Ư(13) =$ {$±1; ±13$}
$⇒ n ∈$ {$-1; -3; 11; -15$}
$Vậy$ $n ∈$ {$-1; -3; 11; -15$} $thì$ $5n-3$ $chia$ $hết$ $cho$ $n+2$
$b) Đặt$ $B = \frac{6n +5}{n -2} = 6 + \frac{17}{n -2}$
$Để$ $B ∈ Z ⇔ \frac{17}{n -2} ∈ Z$
$Mà$ $n ∈ Z$
$⇒ n -2 ∈ Ư(17) =$ {$±1; ±17$}
$⇒ n ∈$ {$3; 1; 19; -15$}
$Vậy$ $n ∈$ {$3; 1; 19; -15$} $thì$ $6n +5$ $chia$ $hết$ $cho$ $n-2$
