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a: \(\Leftrightarrow n-5\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{6;4;8;2\right\}\)
b: \(\Leftrightarrow3n-18+18⋮n-6\)
\(\Leftrightarrow n-6\in\left\{1;-1;2;-2;3;-3;6;-6;9;-9;18;-18\right\}\)
hay \(n\in\left\{7;5;8;4;9;3;12;0;15;-3;24;-12\right\}\)
a) \(-3⋮n-5\)
\(\Rightarrow n-5\inƯ\left(-3\right)=\left\{-1;1;-3;3\right\}\)
Có bảng sau:
n-5 | -1 | 1 | -3 | 3 |
n | 4 | 6 | 2 | 8 |
Vậy...
b)
\(\begin{matrix}3n⋮n-6\\n-6⋮n-6\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}3n⋮n-6\\3n-18⋮n-6\end{matrix}\right.\){18\(⋮\) n-6
\(\Leftrightarrow n-6\inƯ\left(18\right)=\left\{1,2,3,6,9,18,-1,-2,-3,-6,-9,-18\right\}\)
Có bảng sau:
n-6 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 | 9 | -9 | 18 | -18 |
n | 7 | 5 | 8 | 4 | 9 | 3 | 12 | 0 | 15 | -3 | 24 | -12 |
Vậy...
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\(a,\Rightarrow n-1+7⋮n-1\)
Mà \(n-1⋮n-1\Rightarrow7⋮n-1\)
\(\Rightarrow n-1\inƯ\left(7\right)=\left\{1;7\right\}\\ \Rightarrow n\in\left\{2;8\right\}\)
\(b,\Rightarrow3\left(n+1\right)+2⋮n+1\)
Mà \(3\left(n+1\right)⋮n+1\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\inƯ\left(2\right)=\left\{1;2\right\}\\ \Rightarrow n=1\left(n\ne0\right)\)
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A = 6 + 62 + 63 + 64 + ... + 62016
6A = 62 + 63 + 64 + 65 + ... + 62017
6A - A = (62 + 63 + 64 + 65 + ... + 62017) - (6 + 62 + 63 + 64 + ... + 62016)
5A = 62017 - 6
6n = 5A + 6
6n = (62017 - 6) + 6
6n = 62017
=>n = 2017
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a: \(\Leftrightarrow n+8-11⋮n+8\)
\(\Leftrightarrow n+8\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{-7;-9;3;-19\right\}\)
b: Đề thiếu rồi bạn
a, \(\dfrac{n-3}{n+8}=\dfrac{n+8-11}{n+8}=1-\dfrac{11}{n+8}\)
\(\Rightarrow n+8\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
n+8 | 1 | -1 | 11 | -11 |
n | -7 | -9 | 3 | -19 |
b, bạn bổ sung đề nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3n+14⋮n+2\)
=>\(3n+6+8⋮n+2\)
=>\(8⋮n+2\)
=>\(n+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(n\in\left\{-1;-3;0;-4;2;-6;6;-10\right\}\)
mà n>=0
nên \(n\in\left\{0;2;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,Ta có:n+2 chia hết cho n-3
=>n-3+5 chia hết cho n-3
Mà n-3 chia hết cho n-3
=>5 chia hết cho n-3
=>n-3\(\in\)Ư(5)={-5,-1,1,5}
=>n\(\in\){-2,2,4,8}
b,Ta có:2n-7 chia hết cho n-1
=>2n-2-5 chia hết cho n-1
=>2(n-1)-5 chia hết cho n-1
Mà 2(n-1) chia hết cho n-1
=>5 chia hết cho n-1
=>n-1\(\in\)Ư(5)={-5,-1,1,5}
=>n\(\in\){-4,0,2,6}