![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5n-8⋮4-n\)
Mà \(-n+4⋮4-n\)
\(\Leftrightarrow\left\{{}\begin{matrix}5n-8⋮4-n\\-5n+20⋮4-n\end{matrix}\right.\)
\(\Leftrightarrow12⋮4-n\)
\(\Leftrightarrow4-n\inƯ\left(12\right)\)
Tự xét tiếp!
Theo đề bài ta có:
5n - 8 \(⋮\) 4 - n
Mà - n \(⋮\) - n
\(\Rightarrow\) \(\left\{{}\begin{matrix}5n-8⋮4-n\\-5n+20⋮4-n\end{matrix}\right.\)
\(\Leftrightarrow\) 12 \(⋮\) 4 - n
\(\Leftrightarrow\) 4 - n \(\in\) Ư(12)
\(\Leftrightarrow\) 4 - n \(\in\) { 1;2;3;4;6;12 }
\(\Leftrightarrow\) n \(\in\){ 5;6;7;8;10;16}
Vậy n \(\in\) { 5;6;7;8;10;16 }
![](https://rs.olm.vn/images/avt/0.png?1311)
a) n+2 \(\in\)B(3)={0;3;6;9;12;15;18;21;...}
\(\Rightarrow\)n=1;4;7;10;13;16;19;....
b) 4n-5 \(\in\)B(13)={0;13;26;39;42;.....}
\(\Rightarrow\)n=5;18;31;44;47;...
c) 5n-1 \(\in\)B(7)={0;7;14;21;28;35;42;...}
\(\Rightarrow\)n=3
d) 25n+3 \(\in\)B(57)={0;57;114;171;228;285...}
\(\Rightarrow\)n=9
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:5n+3 :n+1
3 chia hết cho n suy ra 3 chia hết cho 1
vậy Ư(3)={1;3}
Vậy n thuộc {1;3}
Ta có:5n+3=5n+5-2=5(n+1)-2
Để 5n+3 chia hết cho n+1 thì 2 chia hết cho n+1
=>n+1\(\in\)U(2)={1,2}
=>n\(\in\){0,1}
tick nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
5n+45 ⋮ n+3
=> [(5n+15)-15+45] ⋮ n+3
=> [(5n+3.5)+30] ⋮ n+3
=> [5.(n+3)+30] ⋮ n+3
có n+3 ⋮ n+3 => 5.(n+3) ⋮ n+3
=> 30 ⋮ n+3
=> n+3 ∈ Ư(30)
=> n+3 ∈ {1;2;3;5;6;10;15;30} mà n ∈ N
=> N ∈ {0;2;3;7;12;27}
vậy_____