Question

Tìm STN n để :

a) n+9\(⋮\)n+4

b)3n+40\(⋮\)n+4

c)5n+2\(⋮\)2n+9

Answer 1

a/ \(n+9⋮n+4\)

Mà \(n+4⋮n+4\)

\(\Leftrightarrow5⋮n+4\)

Vì \(n\in N\Leftrightarrow n+4\in N;n+4\inƯ\left(5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n+4=1\\n+4=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-3\left(loại\right)\\n=1\left(tm\right)\end{matrix}\right.\)

Vậy ...

b/ \(3n+40⋮n+4\)

Mà \(n+4⋮n+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}3n+40⋮n+4\\3n+12⋮n+4\end{matrix}\right.\)

\(\Leftrightarrow28⋮n+4\)

Vì \(n\in N\Leftrightarrow n+4\in N;n+4\inƯ\left(28\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n+4=1\\n+4=28\\n+4=2\\n+4=14\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-3\left(loại\right)\\n=24\\n=-2\left(loại\right)\\n=10\end{matrix}\right.\)

Vậy ...

c/ \(5n+2⋮2n+9\)

Mà \(2n+9⋮2n+9\)

\(\Leftrightarrow\left\{{}\begin{matrix}10n+4⋮2n+9\\10n+45⋮2n+9\end{matrix}\right.\)

\(\Leftrightarrow41⋮2n+9\)

\(\Leftrightarrow2n+9\inƯ\left(41\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n+9=1\\2n+9=41\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-4\left(loại\right)\\n=16\end{matrix}\right.\)

Vậy ..

Answer 2

b)3n+40$⋮$n+4

=> vì \(\left(n+4\right)⋮\left(n+4\right)\)

=> \(\left(3n+12\right)⋮\left(n+4\right)\)

=> \(\left(3n+40-4x-12\right)⋮\left(n+4\right)\)

=> \(28⋮\left(n+4\right)\)

=> \(n+4\inƯ\left(28\right)=\left\{1;2;4;7;14;28\right\}\)

ta có bảng sau

n+4	1	2	4	7	14	28
n	-3 loại	-2loại	0	3	10	24

vậy \(n\in\left\{0;3;10;24\right\}\)