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\(2n+9⋮3n+1\)
\(\Rightarrow3\left(2n+9\right)⋮3n+1\)
\(\Rightarrow2\left(3n+1\right)+25⋮3n+1\)
\(\Rightarrow25⋮3n+1\)
\(\Rightarrow3n+1\in\left\{5,25,1,-5,-25,-1\right\}\)
\(n\in\left\{8,0\right\}\)
\(5n+2⋮9-2n\)
\(\Rightarrow2\left(5n+2\right)⋮9-2n\)
\(\Rightarrow-5\left(9-2n\right)-41⋮9-2n\)
\(41⋮9-2n\)
\(\Rightarrow9-2n\in\left\{41,-41,1,-1\right\}\)
\(\Rightarrow n\in\left\{-16,25,4,-5\right\}\)
a) \(n^2-3n+9\)chia het cho \(n-2\)
\(\Leftrightarrow\)\(n^2-2n-n-2+11\)chia het cho \(n-2\)
\(\Leftrightarrow\)\(\left(n-2\right)\left(n+1\right)+11\)chia het cho \(n-2\)
\(\Leftrightarrow\)11 chia het cho \(n-2\)
\(\Rightarrow\)\(n-2\in U\left(11\right)\)\(\Rightarrow\)\(n-2\in\left\{-11;-1;1;11\right\}\)
\(\Rightarrow\)\(n\in\left\{-9;1;3;13\right\}\)
b) 2n-1 chia hết cho n-2
\(\Rightarrow2n-2+3\) chia hết cho\(n-2\)
\(\Rightarrow3\)chia hết cho \(n-2\)
\(\Rightarrow n-2\in U\left(3\right)\)\(\Rightarrow n-2\in\left\{-3;-1;1;3\right\}\)\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)
a) Vì 3\(⋮\)n
=> n\(\in\)Ư(3)={ 1; 3 }
Vậy, n=1 hoặc n=3
\(------huongdan-----\)
\(Taco:\)
\(\left(3n-2n\right)⋮n+1\Leftrightarrow n⋮n+1\Leftrightarrow\left(n+1\right)-n⋮n+1\Leftrightarrow1⋮n+1\)
\(\Leftrightarrow n+1\in\left\{-1;1\right\}\Leftrightarrow n\in\left\{-2;0\right\}\)
\(b,2n-4⋮n+2\Leftrightarrow2n+4-2n+4⋮2n+4\Leftrightarrow8⋮2n+4\)
dễ thấy: 2n+4 chẵn => 2n+4 là ước chẵn của 8
\(\Rightarrow2n+4\in\left\{2;4;8;-2;-4;-8\right\}\Rightarrow2n\in\left\{-2;0;4;-6;-8;-12\right\}\)
\(\Rightarrow n\in\left\{-1;0;2;-3;-4;-6\right\}\)
b)
Để \(2n⋮\left(n-1\right)\)
\(\Rightarrow2.\left(n-1\right)+2⋮\left(n-1\right)\)
\(\Rightarrow2⋮\left(n-1\right)\)
\(\Rightarrow\left(n-1\right)\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\end{matrix}\right.\)
Vậy n=2;n=3 thì \(2n⋮\left(n-1\right)\)
c)
Để \(\left(3n-8\right)⋮\left(n-4\right)\)
\(\Rightarrow3.\left(n-4\right)+4⋮\left(n-4\right)\)
\(\Rightarrow4⋮\left(n-4\right)\)
\(\Rightarrow\left(n-4\right)\inƯ\left(4\right)=\left\{1;2;4\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}n-4=1\Rightarrow n=5\\n-4=2\Rightarrow n=6\\n-4=4\Rightarrow n=8\end{matrix}\right.\)
Vậy với .....................
2.(3n-4)=6n-8 chia hết (2n-1)
6n-8=3(2n-1)-5 chia hết (2n-1)
2n-1 chia hết cho (2n-1) hiển nhiên
=> 5 phải chia hết cho (2n-1)
2n-1 = ước (5) =(-5,-1,1,5)
2n=(-4,0,2,6)
n={-2,0,1,3}