Question

Tìm số nguyên n biết :

a) \(n+7\) \(⋮\) \(n+2\) e) \(2n+7\) \(⋮\) \(n+1\)

b) \(9-n\) \(⋮\) \(n-3\) h) \(3n+7\) \(⋮\) \(2n+1\)

c) \(n^2+n+17\) \(⋮\) \(n+1\) g) \(3n^2+5\) \(⋮\) \(n-1\)

d) \(n^2+25\) \(⋮\) \(n+2\) i) \(2n^2+11\) \(⋮\) \(3n+1\)

Answer 1

a)\(n+7⋮n+2\)

\(\Rightarrow\left(n+2\right)+5⋮n+2\)

\(\Rightarrow5⋮n+2\)

\(\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau:

n+2	1	-1	5	-5
n	-1	-3	3	-7

Vậy \(n\in\left\{-1;-3;3;-7\right\}\)

b)\(9-n⋮n-3\)

\(\Rightarrow6-\left(n-3\right)\)

\(\Rightarrow6⋮n-3\)

\(\Rightarrow n-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

nếu n-3=1 thì n=4

nếu n-3=-1 thì n=2

nếu n-3=2 thì n=5

nếu n-3=-2 thì n=1

nếu n-3=3 thì n=6

nếu n-3=-3 thì n=0

nếu n-3=6 thì n=9

nếu n-3=-6 thì n=-3

Vậy \(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)

Answer 2

c)\(n^2+n+17⋮n+1\)

\(\Rightarrow n\left(n+1\right)+17⋮n+1\)

\(\Rightarrow17⋮n+1\)

\(\Rightarrow n+1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)

nếu n+1=1 thì n=0

nếu n+1=-1 thì n=-2

nếu n+1=17 thì n=16

nếu n+1=-17 thì n=-18

Vậy \(n\in\left\{0;-2;16;-18\right\}\)