Ta có : \(\frac{A}{B}=\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=\frac{n+2}{n+2}+\frac{3}{n+2}=1+\frac{3}{n+2}\)
\(\Rightarrow n+2\inƯ\left(3\right)=\left\{1,3\right\}\)
Ta có bảng :
n+2 | 1 | 3 |
n | -1 | 1 |
Vậy n = 1
A/B=n+5/n+2=n+2+3/n+2=(n+2/n+2)+(3/n+2)=>3 chia hết cho n+2=>n+2 thuộc Ư(3)={+-1;+-3}
n+2=1=>n=-1
n+2=-1=>n=-3
n+2=3=>n=1
n+2=-3=>n=-5