Ta có \(\left(n+5\right)\left(n+6\right)⋮6n\Rightarrow\dfrac{\left(n+5\right)\left(n+6\right)}{6n}\in Z\)
Ta có \(\dfrac{\left(n+5\right)\left(n+6\right)}{6n}=\dfrac{n^2+11n+30}{6n}=\dfrac{1}{6}\left(n+11+\dfrac{30}{n}\right)\)
Vậy để \(\dfrac{\left(n+5\right)\left(n+6\right)}{6n}\in Z\) thì \(n\inƯ\left(30\right)\in\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\right\}\)
Thử lại: ta có 1;-2;3;-5;-6;10;-15;30 thõa mãn
Vậy n={1;-2;3;-5;-6;10;-15;30} thì P=(n+5)(n+6) chia hết cho 6n