\(a,n+3⋮n\)

mà \(n⋮n\Rightarrow n\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(b,2n+3⋮n\)

mà \(2n⋮n\Rightarrow n\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(c,3n-1⋮n+1\)

\(\Rightarrow3n+3-2⋮n+1\)

\(\Rightarrow3\left(n+1\right)-2⋮n+1\)

\(\Rightarrow n+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow n\in\left\{0;-2;1;-3\right\}\)

a) Ta có : n+ 3 chia hết cho n

           => 3 chia hết cho n

           => n thuộc ước của 3

           => n thuộc {-1;1;3;-3}

b) Ta có: 2n + 3 chia hết cho n

        => 3 chia hết cho n 

        =>n thuộc ước của 3

        => n thuộc {-1;1;3;-3}

c) Ta có : 3n-1 chia hết cho n+1

        =>  3n +3 -4 chia hết cho n+ 1

Mà : 3n + 3 chia hết cho n+1

        => 4 chia hết cho n+1

        => n+1 thuộc ước của 4

        => n+1 thuộc {1;-1;2;-2;4;-4}

        => n thuộc {0;-2;1;-3;3;-5}

d) Mk chưa làm được ! Sorry nha !!

Hok tốt :))

11,

a, 4x-3\(\vdots\) x-2 ¹

    x-2\(\vdots\) x-2\(\Rightarrow\) 4(x-2)\(\vdots\) x-2\(\Rightarrow\) 4x-8\(\vdots\) x-2 ²

Từ ¹ và ² ta có:

(4x-3)-(4x-8)\(\vdots\) x-2

\(\Rightarrow\) 4x-3-4x+8\(\vdots\) x-2

\(\Rightarrow\)       5       \(\vdots\) x-2

\(\Rightarrow\) x-2\(\in\) Ư(5)

\(\Rightarrow\) x-2\(\in\){-5;-1;1;5}

\(\Rightarrow\) x\(\in\) {-3;1;3;7}

Vậy......

Phần b và c làm tương tự như phần a pn nhé!

a) ta có: n+2 chia hết cho n-3

=>(n-3)+5 chia hết cho n-3

Mà n-3 chia hết cho n-3

=>5 chia hết cho n-3

=> n-3 thuộc Ư(5)={1;5;-1;-5}

=> n thuộc {4;8;2;-2}

b) Ta có: 6n+1 chia hết cho 3n-1

=>(6n-2)+2+1 chia hết cho 3n-1

=>2(3n-1) +3 chia hết cho 3n-1

Mà 2(3n-1) chia hết cho 3n-1

=> 3 chia hết cho 3n-1

=> 3n-1 thuộc Ư(3)={1;3;-1;-3}

=> 3n thuộc {2;4;0;-2}

=>n thuộc {2/3 ; 4/3 ; 0 ; -2/3}

Mà n thuộc Z

=>n=0

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