Question

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Answer 1

\(\dfrac{3n+2}{n-3}=\dfrac{3\left(n-3\right)}{n-3}+\dfrac{8}{n-3}=3+\dfrac{8}{n-3}\)

Để \(\dfrac{3n+2}{n-3}\in Z\) thì \(\left(n-3\right)\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

Vì \(n\in N\Rightarrow n\in\left\{4;2;5;1;7;11\right\}\)

Answer 2

a) Để  \(\dfrac{3n+2}{n-3}\) thì \(3n+2⋮n-3\\ \Rightarrow\left(3n-6\right)+4⋮n-3\\ \Rightarrow3\left(n-3\right)+4⋮n-3\)

Mà \(3\left(n-3\right)⋮n-3\Rightarrow4⋮n-3\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\Rightarrow x=\left\{-1;1;2;4;5;7\right\}\)

Mà \(n\in N\Rightarrow n=\left\{1;2;4;5;7\right\}\)

Answer 3

a) \(\dfrac{3n+2}{n-3}\in Z\Rightarrow3n+2⋮n-3\\ \Rightarrow\left(3n-9\right)+11⋮n-3\\ \Rightarrow3\left(n-3\right)+11⋮n-3\)

Mà \(3\left(n-3\right)⋮n-3\Rightarrow11⋮n-3\Rightarrow n-3\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\Rightarrow n=\left\{2;4;-8;14\right\}\)

Mà \(n\in N\Rightarrow n=\left\{2;4;14\right\}\)

Mik làm lại nhé

Answer 4

a: Để \(\dfrac{3n+2}{n-3}\) là số nguyên thì \(3n+2⋮n-3\)

\(\Leftrightarrow11⋮n-3\)

\(\Leftrightarrow n-3\in\left\{1;-1;11;-11\right\}\)

hay \(n\in\left\{4;2;14;-8\right\}\)