\(2\left(n+1\right)-5⋮n-1\Leftrightarrow-5⋮n-1\)
\(\Rightarrow n-1\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)
| n - 1 | 1 | -1 | 5 | -5 |
| n | 2 | 0 | 6 | -4 |
Ta có: 2n-3=2n+2-5=2(n+1)-5 vậy (2n-3)⋮(n+1)⇔5⋮ (n+1)⇔n+1 ϵ Ư(5)⇔n+1 ϵ { -5; -1; 1;5} ⇔ n ϵ {-6; -2; 0; 4}
\(ĐKXĐ:n+1\ne0\Leftrightarrow n\ne-1\)
\(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2\left(n+1\right)-5}{n+1}=2-\dfrac{5}{n+1}\)
Để \(2n-3⋮n+1\Leftrightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow n=\left\{-6;-2;0;4\right\}\)
