+)Ta có :
$\dfrac{3n+9}{n-2}=\dfrac{3(n-2)+15}{n-2}=\dfrac{3(n-2)}{n-2}+\dfrac{15}{n-2}=3+\dfrac{15}{n-2}$
Để $3n+9\vdots n-2$
$⇔15\vdots n-2$
$⇒n-2∈Ư(15)=\{±1;±3;±5;±15\}$
Ta có bảng : (Ảnh 1)
+)Ta có :
$\dfrac{4n-5}{2n+1}=\dfrac{2(2n+1)-7}{2n+1}=\dfrac{2(2n+1)}{2n+1}+\dfrac{-7}{2n+1}$
Để $4n-5\vdots 2n+1$
$⇔-7\vdots 2n+1$
$⇒2n+1∈Ư{-7}=\{±1;±7\}$
Ta có bảng : (Ảnh 2)
+)Ta có :
$\dfrac{6n-3}{2n+2}=\dfrac{3(2n+2)-9}{2n+2}=\dfrac{3(2n+2)}{2n+2}+\dfrac{-9}{2n+2}$
Để $6n-3\vdots 2n+2$
$⇔-9\vdots 2n+2$
$⇒2n+2∈Ư{-9}=\{±1;±3;±9\}$
Ta có bảng : (Ảnh 3)
