Question

Tìm số nguyên n để: -12/n; 15/n-2 và 8/n+1 đồng thời có giá trị nguyên.

Answer 1

\(-\dfrac{12}{n}\in Z\Leftrightarrow n\inƯ_{12}=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\left(1\right)\)

\(\dfrac{15}{n-2}\in Z\Leftrightarrow n-2\inƯ_{15}=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

\(\Leftrightarrow n=\left\{-13;-3;-1;1;3;5;7;17\right\}\left(2\right)\)

\(\dfrac{8}{n+1}\in Z\Leftrightarrow n+1\inƯ_8=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n=\left\{-7;-5;-3;-2;0;1;3;7\right\}\left(3\right)\)

Đến đây lấy tập hợp giá trị của n là giao của \(\left(1\right);\left(2\right);\left(3\right)\)

Answer 2

Để \(-\dfrac{12}{n}\) có giá trị nguyên thì \(12⋮n\)

\(\Leftrightarrow n\inƯ\left(12\right)\)

\(\Leftrightarrow n\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)(1)

Để \(\dfrac{15}{n-2}\) có giá trị nguyên thì \(15⋮n-2\)

\(\Leftrightarrow n-2\inƯ\left(15\right)\)

\(\Leftrightarrow n-2\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(n\in\left\{3;1;5;-1;7;-3;17;-13\right\}\)(2)

Để \(\dfrac{8}{n+1}\) có giá trị nguyên thì \(8⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(8\right)\)

\(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

hay \(n\in\left\{0;-2;1;-3;3;-5;7;-9\right\}\)(3)

Từ (1), (2) và (3) suy ra \(n\in\left\{1;3;-3\right\}\)

Vậy: \(n\in\left\{1;3;-3\right\}\)

Answer 3

Ta có :

+) \(-\dfrac{12}{n}\in Z\Leftrightarrow n\inƯ\left(-12\right)\) \(\Leftrightarrow n\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\) \(\left(1\right)\)

+) \(\dfrac{15}{n-2}\in Z\Leftrightarrow n-2\inƯ\left(15\right)\) \(\Leftrightarrow n-2\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)

\(\Leftrightarrow n\in\left\{-13;-3;-1;1;3;5;7;17\right\}\) \(\left(2\right)\)

+) \(\dfrac{8}{n+1}\in Z\) \(\Leftrightarrow n+1\inƯ\left(8\right)\) \(\Leftrightarrow n+1\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Leftrightarrow n\in\left\{-9;-5;-3;-2;0;1;3;7\right\}\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\) \(\Leftrightarrow n\in\left\{1;3\right\}\)

Vậy...