Lời giải:
a)
Để \(A=\frac{2n}{n+1}\in\mathbb{Z}\) thì \(2n\vdots n+1\)
\(\Leftrightarrow 2(n+1)-2\vdots n+1\)
\(\Leftrightarrow 2\vdots n+1\)
\(\Rightarrow n+1\in\left\{\pm 1;\pm 2\right\}\)
\(\Rightarrow n\in\left\{-2;0; -3;1\right\}\)
b)
Để \(B=\frac{n+4}{2-n}\in\mathbb{Z}\) thì \(n+4\vdots 2-n\)
\(\Leftrightarrow 6-(2-n)\vdots 2-n\)
\(\Leftrightarrow 6\vdots 2-n\)
\(\Rightarrow 2-n\in\left\{\pm 1;\pm 2;\pm 3; \pm 6\right\}\)
\(\Rightarrow n\in\left\{3;1;4; 0; 5;-1; 8; -4\right\}\)
Thử lại thấy đúng. Vậy...........
Lời giải:
a)
Để \(A=\frac{2n}{n+1}\in\mathbb{Z}\) thì \(2n\vdots n+1\)
\(\Leftrightarrow 2(n+1)-2\vdots n+1\)
\(\Leftrightarrow 2\vdots n+1\)
\(\Rightarrow n+1\in\left\{\pm 1;\pm 2\right\}\)
\(\Rightarrow n\in\left\{-2;0; -3;1\right\}\)
b)
Để \(B=\frac{n+4}{2-n}\in\mathbb{Z}\) thì \(n+4\vdots 2-n\)
\(\Leftrightarrow 6-(2-n)\vdots 2-n\)
\(\Leftrightarrow 6\vdots 2-n\)
\(\Rightarrow 2-n\in\left\{\pm 1;\pm 2;\pm 3; \pm 6\right\}\)
\(\Rightarrow n\in\left\{3;1;4; 0; 5;-1; 8; -4\right\}\)
Thử lại thấy đúng. Vậy...........
