a) $\dfrac{n-2}{n+3}$

= n-2 $\vdots$ n+3

= (n+3-5) $\vdots$ n+3

Vì n+3 $\vdots$ n+3 nên n+3 $\in$ Ư(5)={1; -1; 5; -5}

+) n+3=1

=>n=-2 

+)n+3=-1

=>n=-4

+)n+3=5

=>n=2

+)n+3=-5

=>n=-8

Vậy n$\in$ {-2; -4; 2; -8}

b) $\dfrac{2n+1}{n+2}$

2n+1 $\vdots$ n+2

(2n+2)-1 $\vdots$ n+2 

Vì 2n+2 $\vdots$ n+2 nên n+2 $\in$ Ư(-1)= {1; -1}

+) n+2=-1

=>n=-3

+)n+2=1

=>n=-1

Vậy n $\in$ {-3; -1}

d)$\dfrac{3n+2}{n+1}$

3n+2 $\vdots$ n+1 

3(n-1)+5 $\vdots$ n+1

3+5 $\vdots$ n+1

5 $\vdots$ n+1 $\in$ 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={0; -1; 4; -6}

e)$\dfrac{n+1}{n-2}$

n+1 $\vdots$ n-2

(n-2)+3 $\vdots$ n-2

Vì n-2 $\vdots$ n-2 nên n-2 $\in$ Ư(3)={1; -1; 3; -3}

+) n-2=1

n= 3

+) n-2=-1

n=1

+) n-2=3

n= 5

+) n-2=-3

n=-1

Vậy n={3; 1; 5; -1}

f) $\dfrac{4n+5}{2n-1}$

4n+5 $\vdots$ 2n-1

2(2n-2)+7 $\vdots$ 2n-1

Vì 2(2n-2) $\vdots$ 2n-1 nên 2n-1 $\in$ Ư(7)={1;-1; 7; -7}

+)2n-1=1

n=1

+)2n-1=-1

n=0

+)2n-1=7

n=4

+2n-1=-7

n=-3

Vậy n $\in$ ={1; 0; 4; -3}