Ta có : 17-2n chia hết cho n+3 hay \(\dfrac{17-2n}{n+3}\in Z\)
Ta lại có:
\(\dfrac{17-2n}{n+3}=\dfrac{-2n+17}{n+3}=\dfrac{-2n+\left(-2\right).3+23}{n+3}=\dfrac{-2.\left(n+3\right)23}{n+3}\)
vì -2.(n+3) chia hết cho n+3 mà để\(\dfrac{17-2n}{n+3}\in Z\) suy ra :\(\dfrac{23}{n+3}\in Z\)
Suy ra: n+3 \(\in\) \(_{Ư_{\left(23\right)}}\) \(\Rightarrow\) n+3\(\in\left\{1;23;-1;-23\right\}\)
\(\Rightarrow n\in\left\{-2;20;-4;-26\right\}\)
vậy n\(\in\left\{-2;20;-4;-26\right\}\)
