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5n + 19 chia hết cho 2n + 1
⇒ 2(5n + 19) chia hết cho 2n + 1
⇒ 10n + 38 chia hết cho 2n + 1
⇒ 10n + 5 + 33 chia hết cho 2n + 1
⇒ 5(2n + 1) + 33 chia hết cho 2n + 1
⇒ 33 chia hết cho 2n + 1
⇒ 2n + 1 ∈ Ư(33) = {1; -1; 3; -3; 11; -11; 33; -33}
Mà: n ∈ N
⇒ 2n + 1 ∈ {1; 3; 11; 33}
⇒ n ∈ {0; 1; 5; 16}
5n+19 chia hết cho 2n+1
=> 10n+38 chia hết cho 2n+1
=> 5(2n+1)+33 chia hết cho 2n+1
=> 33 chia hết cho 2n+1 ( Vì 5(2n+1) luôn chia hết cho 2n+1 với n là STN )
=> 2n+1 thuộc Ư(33)={1;-1;33;-33}
=> 2n thuộc {0;-2;32;-34}
=> n thuộc {0;-1;16;-17}
Đến đây bạn thử lại từng giá trị của x vào đề bài rồi kết luận nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
n+1\(⋮\)7
\(\Rightarrow\)5n+1+14\(⋮7\)
\(\Rightarrow5n+15⋮7\)
\(\Rightarrow5(n+3)⋮7\)
\(\Rightarrow n+3⋮7\left(vi(5:7)=1\right)\)
\(\Rightarrow n+3\in B_{\left(7\right)}\)
\(\Rightarrow n+3=7k\left(k\inℕ^∗\right)\)
\(\Rightarrow n=7k-3\)
vậy n có dạng 7k-3
a, Ta có : 2n + 19 chia hết cho 7
\(\Rightarrow\) \(2n+19\inƯ\left(7\right)\)
Mà \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\Rightarrow\) \(2n+19\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow\) \(2n\in\left\{20;18;26;12\right\}\)
\(\Rightarrow\) \(n\in\left\{10;9;13;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) Ta có : 2n - 2 = 2(n - 1) chia hết cho n - 1
Nên với mọi giá trị của n thì 2n - 2 đều chia hết cho n - 1
3) Ta có : 5n - 1 chia hết chi n - 2
=> 5n - 10 + 9 chia hết chi n - 2
=> 5(n - 2) + 9 chia hết chi n - 2
=> n - 2 thuộc Ư(9) = {1;3;9}
Ta có bảng :
n - 2 | 1 | 3 | 9 |
n | 3 | 5 | 11 |
1) Ta có : 2n + 3 chia hết cho 3n + 1
<=> 6n + 9 chia hết cho 3n + 1
<=> 6n + 2 + 7 chia hết cho 3n + 1
=> 7 chia hết cho 3n + 1
=> 3n + 1 thuộc Ư(7) = {1;7}
Ta có bảng :
3n + 1 | 1 | 7 |
3n | 0 | 6 |
n | 0 | 2 |
Vậy n thuộc {0;2}
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ta có
3n+3 =3(n+1) luôn chia hết cho n+1 với mọi số tự nhiên n
b. ta có :\(5n+19\text{ chia hết cho 2n+1 thì }10n+38\text{ cũng chia hết cho 2n+1}\)
mà \(10n+38=5\left(2n+1\right)+33\text{ chia hết cho }2n+1\) khi 33 chia hết cho 2n+1
hay \(2n+1\in\left\{1,3,11,33\right\}\Rightarrow n\in\left\{0,1,5,16\right\}\)
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a, 2n + 3 ⋮ n ( n \(\ne\) 0)
3 ⋮ n
n \(\in\) Ư(3) = { -3; -1; 1; 3}
b, 2n + 16 ⋮ n + 1 ( n \(\ne\) -1)
2(n + 1) + 14 ⋮ n + 1
14 ⋮ n + 1
n + 1 \(\in\) { -14; -7; -2; -1; 1; 2; 7; 14}
n \(\in\) {-15; - 8; -3; -2; 0; 1; 6; 13}
c, 5n + 12 ⋮ n - 3 (n \(\ne\) 3)
5.(n - 3) + 27 ⋮ n - 3
27 ⋮ n -3
n - 3 \(\in\) {-27; -9; -3; -1; 1; 3; 9; 27}
n \(\in\) {-24; -6; 0; 2; 6; 12; 30}
a) (2n + 3) ⋮ n khi 3 ⋮ n
⇒ n ∈ {-3; -1; 1; 3}
b) 2n + 16 = 2n + 2 + 14 = 2(n + 1) + 14
Để (2n + 16) ⋮ (n + 1) thì 14 ⋮ (n + 1)
⇒ n + 1 ∈ Ư(14) = {-14; -7; -2; -1; 1; 2; 7; 14}
⇒ n ∈ {-15; -8; -3; -2; 0; 1; 6; 13}
c) Ta có:
5n + 12 = 5n - 15 + 27 = 5(n - 3) + 27
Để (5n + 12) ⋮ (n - 3) thì 27 ⋮ (n - 3)
⇒ n - 3 ∈ Ư(27) = {-27; -9; -3; -1; 1; 3; 9; 27}
⇒ n ∈ {-24; -6; 0; 2; 4; 6; 12; 30}
tk:
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