\(a^2+a=a\left(a+1\right)\)
Để \(a\left(a+1\right)⋮\left(a-3\right)\) thì \(\left[\begin{matrix}\frac{a+1}{a-3}\in Z\\\frac{a}{a-3}\in Z\end{matrix}\right.\)
\(\frac{a+1}{a-3}=\frac{a-3+4}{a-3}=1+\frac{4}{a-3}\Rightarrow a-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow a\in\left\{2;4;1;5;-1;7\right\}\)
\(\frac{a}{a-3}=\frac{a-3+3}{a-3}=1+\frac{3}{a-3}\Rightarrow a-3\in\left\{\pm1;\pm3\right\}\)
\(\Rightarrow a\in\left\{2;4;0;6\right\}\)
Vậy để \(\left(a^2+a\right)⋮\left(a-3\right)\) thì \(a\in\left\{\pm1;0;2;4;5;6;7\right\}\)
\(a^2+a⋮a-3\)
\(\Rightarrow a^2-3a+3a+a⋮a-3\)
\(\Rightarrow a\left(a-3\right)+4a⋮a-3\)
\(\Rightarrow4a⋮a-3\)
\(\Rightarrow4a-12+12⋮a-3\)
\(\Rightarrow4\left(a-3\right)+12⋮a-3\)
\(\Rightarrow12⋮a-3\)
\(\Rightarrow a-3\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
| a - 3 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
| a | 4 | 2 | 5 | 1 | 6 | 0 | 7 | -1 | 9 | -3 | 15 | -9 |
Vậy a = {4;2;5;1;6;0;7;-1;9;-3;15;-9}
