A= (n-5)/(n+1) = (n+1-6)/(n+1) = (n+1)/(n+1) - 6/(n+1) = 1-6/(n+1)
để A thuộc Z thì n+1 thuộc Ư(6)...
Ta có:
\(\frac{n-5}{n+1}=\frac{\left(n+1\right)-4}{n+1}=\frac{n+1}{n+1}-\frac{4}{n+1}=1-\frac{4}{n+1}\)
Để A \(\in\) Z thì \(\frac{4}{n+1}\in Z\)
\(\Rightarrow\) 4 chia hết cho n + 1
\(\Rightarrow n+1\inƯ_{\left(4\right)}\)
\(\Rightarrow n+1\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow n\in\left\{0;1;3;-2;-3;-5\right\}\)
Vậy \(n\in\left\{0;1;3;-2;-3;-5\right\}\)
Ai k mình, mình k lại.
Ta có:
\(\frac{n-5}{n+1}=\frac{\left(n+1\right)-6}{n+1}=1-\frac{6}{n+1}\)
\(A\in Z\Leftrightarrow\frac{6}{n+1}\in Z\)
\(\Rightarrow6\) chai hết cho n + 1
\(\Rightarrow n+1\inƯ_{\left(6\right)}\)
\(\Rightarrow n+1\left\{1;2;3;6;-1;-2;-3;-6\right\}\)
\(\Rightarrow n\in\left\{0;1;2;5;-2;-3;-4;-7\right\}\)
Vậy \(n\in\left\{0;1;2;5;-2;-3;-4;-7\right\}\)