a) Ta có:
\(2n+1⋮n-3\)
\(\Rightarrow\left(2n-6\right)+7⋮n-3\)
\(\Rightarrow2\left(n-3\right)+7⋮n-3\)
\(\Rightarrow7⋮n-3\)
\(\Rightarrow n-3\in\left\{1;7\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-3=1\Rightarrow n=4\\n-3=7\Rightarrow n=10\end{matrix}\right.\)
Vậy n=4 hoặc n=10
b) Ta có:
\(n^2+3n-13⋮n+3\)
\(\Rightarrow n\left(n+3\right)-13⋮n+3\)
\(\Rightarrow-13⋮n+3\)
\(\Rightarrow n+3\in\left\{1;13\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n+3=1\Rightarrow n=-2\left(loai\right)\\n+3=13\Rightarrow n=10\end{matrix}\right.\)
Vậy n=10
c) Ta có:
\(n^2+3⋮n-1\)
\(\Rightarrow n^2-1+4⋮n-1\)
\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
\(\Rightarrow n+1+4⋮n-1\)
\(\Rightarrow n+5⋮n-1\)
\(\Rightarrow\left(n-1\right)+6⋮n-1\)
\(\Rightarrow6⋮n-1\)
\(\Rightarrow n-1\in\left\{1;2;3;6\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\\n-1=3\Rightarrow n=4\\n-1=6\Rightarrow n=7\end{matrix}\right.\)
Vậy n=2 hoặc n=3 hoặc n=4 hoặc n=7
a,\(2n+1=2n-6+7=2\left(n-3\right)+7\)
Do \(2\left(n-3\right)⋮n-3\)
\(\Rightarrow n-3\in\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}n-3=1\\n-3=-1\\n-3=7\\n-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=4\\n=2\\n=10\\n=-4\end{matrix}\right.\)
(2n+1) ⋮ (n-3)
vì (n-3)⋮(n-3)
➩2(n-3)⋮(n-3)
➩(2n-6)⋮(n-3)
➩(2n+1)-(2n-6)⋮(n-3)
➩(2n+1-2n+6)⋮(n-3)
➩7⋮(n-3)
➩(n-3)∈Ư(7)= {1;7}
ta có bảng sau
| n-3 | 1 | 7 |
| n | 4 | 10 |
vậy n=4 hoặc n=10