=> \(\dfrac{2n^2+n-7}{n-2}=2n+5+\dfrac{3}{n-2}\)
để 2n2+n-7 ⋮n-2 thì 2n+5+\(\dfrac{3}{n-2}\) nguyên
=> \(\dfrac{3}{n-2}\) nguyên
=> 3⋮n-2
=> n-2 ∈Ư(3)=\(\left\{\pm1;\pm3\right\}\)
ta có bảng sau
| n-2 | -3 | -1 | 1 | 3 |
| n | -1 | 1 | 3 | 4 |
| nx | tm | tm | tm | tm |
vậy n∈{-1;1;3;4}
Ta có:
\(2n^2+n-7⋮n-2\)
\(\Rightarrow2n^2+\left(n-2\right)-5⋮n-2\)
\(\Rightarrow2n^2-5⋮n-2\)
\(\Rightarrow\left(2n^2-8\right)+3⋮n-2\)
\(\Rightarrow2\left(n^2-4\right)+3⋮n-2\)
\(\Rightarrow2\left(n-2\right)\left(n+2\right)+3⋮n-2\)
\(\Rightarrow3⋮n-2\)
\(\Rightarrow n-2\in\left\{-1;1;-3;3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n-2=-1\Rightarrow n=1\\n-2=1\Rightarrow n=3\\n-2=-3\Rightarrow n=-1\\n-2=3\Rightarrow n=5\end{matrix}\right.\)
Vậy \(n\in\left\{1;3;-1;5\right\}\)
(2n2+n-7) : (n-2) = n+5 dư 3
Vì 2n2+n-7 chia hết cho n-2
=> \(\left(n-2\right)\in\) Ư(3)=( 1;-1;3;-3)
=> n\(\in\) (3;1;5;-1)
Vậy .....
