Question

Tìm số nguyên n 

1\   n+7 chia hết cho n+2

2\   9-n chia hết cho n-3

3\   n^2+n+17 chia hết cho n+1

4\   n^2+25 chia hết cho n+2

5\   2n+7 chia hết cho n+1

6\   3n+7 chia hết cho 2n+1

Answer 1

ta có : n+7 chia hết n+2

=> (n+2)+5 chia hết cho n+2

=> 5 chia hết n+2

=> n+2 c Ư (5) = { 1;5 }

+) n+2 = 1 => n=-1

+) n+2=5 => n=3

vậy n = -1 và n = 3

Answer 2

Ta có:

\(n+7⋮n+2\)

\(\Leftrightarrow\left(n+2\right)+5⋮n+2\)

Vì \(n+2⋮n+2\)

Để \(\left(n+2\right)+5⋮n+2\)

Thì \(5⋮n+2\)

\(\Rightarrow n+2\inƯ\left(5\right)=\left\{1;5\right\}\)

\(\Rightarrow\orbr{\begin{cases}n+2=1\\n+2=5\end{cases}\Rightarrow\orbr{\begin{cases}n=-1\\n=3\end{cases}}}\)

Vậy....

Answer 3

3,\(n^2+n+17⋮n+1\)

\(=>n.\left(n+1\right)+17⋮n+1\)

Do \(n.\left(n+1\right)⋮n+1\)

\(=>17⋮n+1\)

\(=>n+1\inƯ\left(17\right)\)

\(=>n+1\in\left\{-17;-1;1;17\right\}\)

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

Answer 4

4.\(n^2+25⋮n+2\)

\(=>\left(n+5\right)^2⋮n+2\)

\(=>n+5⋮n+2\)

\(=>3⋮n+2\)

\(=>n+2\inƯ\left(3\right)\)

\(=>n+2\in\left(-1;-3;1;3\right)\)

\(=>n\in\left\{-3;-5;-1;1\right\}\)

Answer 5

5,\(2n+7⋮n+1\)

\(=>2.\left(n+1\right)+5⋮n+1\)

\(=>5⋮n+1\)

\(=>n+1\inƯ\left(5\right)\)

\(=>n+1\in\left\{-1;-5;1;5\right\}\)

\(=>n\in\left\{-2;-6;0;4\right\}\)

Answer 6

6,\(3n+7⋮2n+1\)

\(=>\left(2n+1\right)+\left(n+6\right)⋮2n+1\)

\(=>n+6⋮2n+1\)

\(=>6⋮n+1\)

\(=>n+1\inƯ\left(6\right)\)

\(=>n+1\in\left\{-1;-6;1;6\right\}\)

\(=>n\in\left\{-2;-7;0;5\right\}\)