Question

Tìm số nguyên n, sao cho n^2 + 3n - 13 chia hết cho n+3 .

Answer 1

\(T=\dfrac{n^2+3n-13}{n+3}=\dfrac{n^2-9+3n+9-13}{n+3}\)

\(T=\dfrac{n^2-9}{n+3}+\dfrac{3n+9}{n+3}-\dfrac{13}{n+3}\)

\(T=\dfrac{\left(n+3\right)\left(n-3\right)}{n+3}+\dfrac{3\left(n+3\right)}{n+3}-\dfrac{13}{n+3}\)

\(T=n-3+3-\dfrac{13}{n+3}=n-\dfrac{13}{n+3}\)

\(\Rightarrow13⋮n+3\Leftrightarrow n+3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

\(\Rightarrow\left[{}\begin{matrix}n+3=1\\n+3=-1\\n+3=13\\n+3=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=-4\\n=10\\n=-16\end{matrix}\right.\)