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\(Taco:\)
\(\left(3n-2n\right)⋮n+1\Leftrightarrow n⋮n+1\Leftrightarrow\left(n+1\right)-n⋮n+1\Leftrightarrow1⋮n+1\)
\(\Leftrightarrow n+1\in\left\{-1;1\right\}\Leftrightarrow n\in\left\{-2;0\right\}\)
\(b,2n-4⋮n+2\Leftrightarrow2n+4-2n+4⋮2n+4\Leftrightarrow8⋮2n+4\)
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 .....................
a) (2n-1)4 : (2n-1) = 27
(2n-1)3 = 27 =33
=> 2n - 1= 3
=> 2n = 4
n = 2
phần b,c làm tương tự nha bn
d) (21+n) : 9 = 95:94
(2n+1) : 9 = 9
2n + 1 = 81
2n = 80
n = 40
a) Vì 3\(⋮\)n
=> n\(\in\)Ư(3)={ 1; 3 }
Vậy, n=1 hoặc n=3
\(2n+9⋮3n+1\)
\(\Rightarrow3\left(2n+9\right)⋮3n+1\)
\(\Rightarrow2\left(3n+1\right)+25⋮3n+1\)
\(\Rightarrow25⋮3n+1\)
\(\Rightarrow3n+1\in\left\{5,25,1,-5,-25,-1\right\}\)
\(n\in\left\{8,0\right\}\)
\(5n+2⋮9-2n\)
\(\Rightarrow2\left(5n+2\right)⋮9-2n\)
\(\Rightarrow-5\left(9-2n\right)-41⋮9-2n\)
\(41⋮9-2n\)
\(\Rightarrow9-2n\in\left\{41,-41,1,-1\right\}\)
\(\Rightarrow n\in\left\{-16,25,4,-5\right\}\)
a)
\(n+5⋮n+1\)
\(\Rightarrow n+1+4⋮n+1\)
\(\Rightarrow4⋮n+1\Rightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{0;-2;1;-3;3;-5\right\}\)
\(a,\left(n+5\right)⋮\left(n+1\right)\Leftrightarrow\left(n+1\right)+4⋮\left(n+1\right)\)
\(\Leftrightarrow4⋮n+1\left(n\inℤ\right)\)
\(\Leftrightarrow n+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow n=-2;0;-3;1;-5;3\)
Vậy \(n=-5;-3;-2;0;1;3\)