a) Ta có: \(n+10⋮n+3\)
\(\Rightarrow\left(n+3\right)+7⋮n+3\)
\(\Rightarrow7⋮n+3\)(vì \(n+3⋮n+3\))
\(\Rightarrow n+3\inƯ\left(7\right)\)
\(\Rightarrow n+3\in\left\{1;7\right\}\)
\(\Rightarrow n=4\)
b)Ta có: \(n-5\inƯ\left(n+8\right)\)
\(\Rightarrow n+8⋮n-5\)
\(\Rightarrow\left(n-5\right)+13⋮n-5\)
\(\Rightarrow n-5\inƯ\left(13\right)\)(vì \(n-5⋮n-5\))
\(\Rightarrow n-5\in\left\{1;13\right\}\)
\(\Rightarrow n\in\left\{6;18\right\}\)
Hok tốt nha^^
a) \(\left(n+10\right)⋮\left(n+3\right)\)
Do: \(\left(n+3\right)⋮\left(n+3\right)\)
Mà: \(\left(n+10\right)⋮\left(n+3\right)\)
\(\Rightarrow\left(n+10\right)-\left(n+3\right)⋮\left(n+3\right)\)
\(\Rightarrow7⋮\left(n+3\right)\rightarrow\left(n+3\right)\varepsilonƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{1;7\right\}\)
Ta có bảng sau:
Mà \(n\varepsilon N\) => n=4
b) Tương tự phần a nha, mk lười viết lm
