Exercice 17 --- (id : 1018)
Activités numériques I: Exercice 17
correction
1) $$\begin{equation*} \begin{split} x&=\dfrac{n+14}{n+3} \\ &=\dfrac{(n+3)+11}{n+3} \\ &=\dfrac{n+3}{n+3}+\dfrac{11}{n+3} \\ &=1+\dfrac{11}{n+3} \end{split} \end{equation*} $$
2) $$\begin{align*} & x\in \Bbb N \\ &\iff \frac{11}{n+3} \in \Bbb N \\ &\iff n+3\; divise\; 11 \\ &\iff n+3=1\; ou \; n+3=11 \\ &\iff n=-2 \notin \Bbb N \; ou \; n=11-3=8 \\ \end{align*}$$ Conclusion: $$n=8$$