\(P=\dfrac{n+13}{n-2}=\dfrac{A}{B}\)
Để P tối giản =>d= (A,B)=1
\(A-B=\left(n+13\right)-\left(n-2\right)=15\)(lớp 6 =>nhiều ước quá)
\(d\le15\)
\(d=2\Rightarrow n=2k\Rightarrow n+13=2k+13̸⋮d\Rightarrow d\ne2k\)
\(d=3\Rightarrow n=3k+2\Rightarrow n+13=3k+15=3\left(k+5\right)⋮3\Rightarrow n\ne3k+2\)
\(d=5\Rightarrow n=5k+2\Rightarrow n+13=5k+15=5\left(k+3\right)⋮5\Rightarrow n\ne5k+2\)
\(d=7\Rightarrow n=7k+2\Rightarrow n+13=\left(7k+15\right)⋮5\Rightarrow d\ne7k\)
\(d=11\Rightarrow n=11k+2\Rightarrow n+13=\left(11k+15\right)⋮̸11\Rightarrow d\ne11k\)
\(d=13\Rightarrow n=13k+2\Rightarrow n+13=\left(13k+15\right)⋮̸13\Rightarrow d\ne13k\)
Kết luận: \(\left\{{}\begin{matrix}n\ne3k+2\\n\ne3k+2\\k\in N,\end{matrix}\right.\)
để p/số \(\dfrac{n+13}{n-2}\) tối giản thì n+13 \(⋮̸\)n-2
=> n+13-(n-2) \(⋮̸\)n-2 hay 15 \(⋮̸\)n-2
=>n-2\(\notin\){1;3;5;15}
=>n\(\notin\){-1;1;3;13}
tick cho mình nha
