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a) \(\Rightarrow\left(n+1\right)+5⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;4\right\}\)
b) \(\Rightarrow2\left(2n+1\right)+7⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;3\right\}\)
Để n + 6 ⋮ n + 1 thì :
⇒ n + 1 + 5 ⋮ n + 1 mà n + 1 ⋮ n + 1
Như thế 5 ⋮ n + 1 và n + 1 ∈ Ư(5)
⇒ Ư(5)={ 1;5 }
n + 1 = 1 ⇒ n = 0
n + 1 = 5 ⇒ n = 4
Vậy .............
⋮⋮⋮![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n+1\in\left\{1;3;9\right\}\)
hay \(n\in\left\{0;1;4\right\}\)
\(a,\Leftrightarrow10n+14⋮2n+1\\ \Leftrightarrow5\left(2n+1\right)+9⋮2n+1\\ \Leftrightarrow2n+1\inƯ\left(9\right)=\left\{1;3;9\right\}\\ \Leftrightarrow n\in\left\{0;1;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{1;0;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow2n-1\in\left\{-1;1;3\right\}\)
hay \(n\in\left\{0;1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(n+6\right)⋮\left(n+1\right)\Rightarrow\left(n+1\right)+5⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;4\right\}\)
b) \(\left(4n+9\right)⋮\left(2n+1\right)\Rightarrow2\left(2n+1\right)+7⋮\left(2n+1\right)\)
\(\Rightarrow\left(2n+1\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{0;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: n+3 chia hết cho n-1
mà: n-1 chia hết cho n-1
suy ra:[(n+3)-(n-1)]chia hết cho n-1
(n+3-n+1)chia hết cho n-1
4 chia hết cho n-1
suy ra n-1 thuộc Ư(4)
Ư(4)={1;2;4}
suy ra n-1 thuộc {1;2;4}
Ta có bảng sau:
n-1 1 2 4
n 2 3 5
Vậy n=2 hoặc n=3 hoặc n=5
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
Ta có: 4n-5 chia hết cho 2n-1
=>4n-2-3 chia hết cho 2n-1
=>2.(2n-1)-3 chia hết cho 2n-1
=>3 chia hết cho 2n-1
=>2n-1=Ư(3)=(-1,-3,1,3)
=>2n=(0,-2,2,4)
=>n=(0,-1,1,2)
Vậy n=0,-1,1,2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
bn cho sai đề rồi
21.2n+1=167
21.2n=167+1
21.2n=168
2n=168:21=8
2n=8=>n=8:2=4
\(a,6⋮n+1\Rightarrow n+1\inƯ\left(6\right)=\left\{1;2;3;6\right\}\Rightarrow n\in\left\{0;1;2;5\right\}\\ b,21⋮2n+1\Rightarrow2n+1\inƯ\left(21\right)=\left\{1;3;7;21\right\}\Rightarrow n\in\left\{0;1;3;10\right\}\)
Cảm ơn nha !