\(A=\dfrac{2n+9}{n+2}-\dfrac{3n}{n+2}+\dfrac{5n+17}{n+2}\)
\(=\dfrac{2\left(n+2\right)+5}{n+2}-\dfrac{3\left(n+2\right)-6}{n+2}+\dfrac{5\left(n+2\right)+7}{n+2}\)
\(=\left(2+\dfrac{5}{n+2}\right)-\left(3-\dfrac{6}{n+2}\right)+\left(5+\dfrac{7}{n+2}\right)\)
\(=2+\dfrac{5}{n+2}-3+\dfrac{6}{n+2}+5+\dfrac{7}{n+2}\)
\(=\left(2-3+5\right)+\left(\dfrac{5}{n+2}+\dfrac{6}{n+2}+\dfrac{7}{n+2}\right)\)
\(=4+\dfrac{5+6+7}{n+2}\)
\(=4+\dfrac{18}{n+2}\)
Để A thuộc Z <=> \(\dfrac{18}{n+2}\in Z\)
<=> 18 chia hết cho n + 2
<=> n + 2 thuộc Ư(18) = {1; 2; 3; 6; 9; 18} (vì n thuộc N)
=> n = -1; 0; 1; 4; 7; 16
Trong các giá trị trên thì chỉ có -1 là không thỏa mãn.
Vậy n = 0; 1; 4; 7; 16
@Đỗ Thị Huyền Trang
\(A=\dfrac{2n+9}{n+2}-\dfrac{3n}{n+2}+\dfrac{5n+17}{n+2}\)
\(=\dfrac{2n+9-3n+5n+17}{n+2}\)
\(=\dfrac{4n+26}{n+2}\)
\(=\dfrac{4n+8+18}{n+2}\)
\(=\dfrac{4\left(n+2\right)+18}{n+2}=4+\dfrac{18}{n+2}\)
Để \(A\in Z\Rightarrow18⋮n+2\)
\(\Rightarrow n+2\in\left\{1;2;3;6;9;18\right\}\) ( do \(n+2\in N\) )
\(\Rightarrow n\in\left\{0;1;4;7;16\right\}\)( do \(n\in N\) )
Vậy ...
