a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)

hay \(n\in\left\{2;4;14\right\}\)

Lời giải:

a.

$3n+2\vdots n-3$

$3(n-3)+11\vdots n-3$

$\Rightarrow 11\vdots n-3$

$\Rightarrow n-3\in\left\{1; -1; 11; -11\right\}$

$\Rightarrow n\in\left\{4; 2; 14; -8\right\}$

Vì $n$ tự nhiên nên $n\in\left\{4;2;14\right\}$

b.

$n^2+7n+9\vdots n+7$

$n(n+7)+9\vdots n+7$

$\Rightarrow 9\vdots n+7$

$\Rightarrow n+7\in\left\{1; -1; 3; -3; 9; -9\right\}$

$\Rightarrow n\in\left\{-6; -8; -4; -10; 2; -16\right\}$

Vì $n$ tự nhiên nên $n=2$

a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)

hay \(n\in\left\{2;4;14\right\}\)

a, 

Ta có: 4n-5 chia hết cho 2n-1

=>4n-2-3 chia hết cho 2n-1

=>2.(2n-1)-3 chia hết cho 2n-1

=>3 chia hết cho 2n-1

=>2n-1=Ư(3)=(-1,-3,1,3)

=>2n=(0,-2,2,4)

=>n=(0,-1,1,2)

Vậy n=0,-1,1,2

b, 

a: 7n chia hết cho 3

mà 7 không chia hết cho 3

nên \(n⋮3\)

=>\(n=3k;k\in Z\)

b: \(-22⋮n\)

=>\(n\inƯ\left(-22\right)\)

=>\(n\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)

c: \(-16⋮n-1\)

=>\(n-1\inƯ\left(-16\right)\)

=>\(n-1\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

=>\(n\in\left\{2;0;3;-1;5;-3;9;-7;17;-15\right\}\)

d: \(n+19⋮18\)

=>\(n+1+18⋮18\)

=>\(n+1⋮18\)

=>\(n+1=18k\left(k\in Z\right)\)

=>\(n=18k-1\left(k\in Z\right)\)

3n+2 chia hết cho n-1

=>3(n-1)+5 chia hết cho n-1

=>5 chia hết cho n-1

=>n-1 E Ư(5)={-1;1;-5;5}

+)n-1=-1=>n=0

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

+)n-1=-5=>n=-4

+)n-1=5=>n=6

vậy...

\(n^2+2n-7:n+2=>n\left(n+2\right)-7:n+2\) ) (: là chia hết)

=>-7 chia hết cho n+2

=>n+2 E Ư(-7)={-1;1;-7;7}

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

+)n+2=1=>n=3

+)n+2=-7=>n=-5

+)n+2=7=>n=9

vậy...

tick nhé

câu a n = 2 là ok

k con khỉ khô

\(\Rightarrow n\left(n+7\right)+9⋮n+7\\ \Rightarrow n+7\inƯ\left(9\right)=\left\{1;3;9\right\}\\ \Rightarrow n=2\left(n\in N\right)\)

\(\Leftrightarrow n+7=9\)

hay n=2

a) 2n+1⋮n-3

2n-6+7⋮n-3

2n-6⋮n-3 ⇒7⋮n-3

n-3∈Ư(7)

Ư(7)={1;-1;7;-7}

⇒n∈{4;2;10;-4}

a. \(2n+7⋮n+1\)

Mà \(n+1⋮n+1\)

\(\Leftrightarrow\hept{\begin{cases}2n+7⋮n+1\\2n+2⋮n+1\end{cases}}\)

\(\Leftrightarrow5⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(5\right)\)

Suy ra :

+) n + 1 = 1 => n = 0

+) n + 1 = 5 => n = 4

Vậy ........