a) 5⋮(n+1)
⇒ (n+1)∈Ư(5)
(n+1)∈Ư(5)={1;5}
⇒n={0;4}
b.
\(n+4⋮n+1\\ \Rightarrow\left(n+1\right)+3⋮n+1\\ \Rightarrow3⋮n+1\)
c.
\(3n+15⋮n+2\\ \Rightarrow3\left(n+2\right)+9⋮n+2\\ \Rightarrow9⋮n+2\)
a)Do 5 ⋮ n+1 (n ϵ N)
\(\Rightarrow n+1\inƯ\left(5\right)\)\(=\left\{\pm1;\pm5\right\}\)
Ta có:
\(n+1=\left(-1\right)\Rightarrow n=\left(-2\right)\)
\(n+1=1\Rightarrow n=0\)
\(n+1=5\Rightarrow n=4\)
\(n+1=\left(-5\right)\Rightarrow n=\left(-6\right)\)
Mà \(n\in N\Rightarrow n=\left\{0;4\right\}\)
b) Từ (n+4) ⋮ (n+1) \(\left(n\in N\right)\)
\(\Rightarrow\left(n+1\right)+3⋮\left(n+1\right)\)
\(\Rightarrow3⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có:
\(\left(n+1\right)=\left(-1\right)\Rightarrow n=\left(-2\right)\)
\(n+1=1\Rightarrow n=0\)
\(n+1=\left(-3\right)\Rightarrow n=\left(-4\right)\)
\(n+1=3\Rightarrow n=2\)
Mà \(n\in N\Rightarrow n=\left\{0,2\right\}\)
c) Từ (3n +15)⋮(n+2) (n ϵ N)
⇒9 + 3(n+2) ⋮ (n+2)
⇒9 ⋮ (n+2)
\(\Rightarrow\left(n+2\right)\inƯ\left(9\right)=\left\{1,3,9\right\}\)
Ta có:
\(n+2=1\Rightarrow n=\left(-1\right)\)
\(n+2=3\Rightarrow n=1\)
\(n+2=9\Rightarrow n=7\)
Mà \(n\in N\Rightarrow n=\left\{1,7\right\}\)
a) 5 \(⋮\) ( n + 1 )
\(\Rightarrow\) n + 1 \(\in\) Ư(5) = { 1;5 }
\(\Rightarrow\) n \(\in\) { 0;4 }
Vậy n \(\in\) { 0;4 }
b) ( n + 4 ) \(⋮\) ( n + 1 )
\(\Rightarrow\) ( n + 1 + 3 ) \(⋮\) ( n + 1 )
Mà ( n + 1 ) \(⋮\) ( n + 1 )
\(\Rightarrow\) 3 \(⋮\) ( n + 1 )
\(\Rightarrow\) n + 1 \(\in\) Ư(3) = { 1;3 }
\(\Rightarrow\) n \(\in\) { 0;2 }
Vậy n \(\in\) { 0;2 }
c) ( 3n + 15 ) \(⋮\) ( n + 2 )
\(\Rightarrow\) 3(n+2) + 9 \(⋮\) ( n + 2 )
Mà 3(n+2) \(⋮\) ( n + 2 )
\(\Rightarrow\) 9 \(⋮\) ( n + 2 )
\(\Rightarrow\) n + 2 \(\in\) Ư(9) = { 1;3;9 }
\(\Rightarrow\) n \(\in\) { -1;1;7 }
Mà n là số tự nhiên
Vậy n \(\in\) { 1;7 }
5 ⋮ ( n + 1 )
⇒ n + 1 ∈ Ư(5) = { 1;5 }
⇒⇒ n ∈ { 0;4 }
Vậy n ∈ { 0;4 }
