tìm số tụ nhiên n để n n2+3n+6 chia hết cho n+3
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4 tháng 9 2023
11:
n^3-n^2+2n+7 chia hết cho n^2+1
=>n^3+n-n^2-1+n+8 chia hết cho n^2+1
=>n+8 chia hết cho n^2+1
=>(n+8)(n-8) chia hết cho n^2+1
=>n^2-64 chia hết cho n^2+1
=>n^2+1-65 chia hết cho n^2+1
=>n^2+1 thuộc Ư(65)
=>n^2+1 thuộc {1;5;13;65}
=>n^2 thuộc {0;4;12;64}
mà n là số tự nhiên
nên n thuộc {0;2;8}
Thử lại, ta sẽ thấy n=8 không thỏa mãn
=>\(n\in\left\{0;2\right\}\)
B
1
26 tháng 12 2021
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
NQ
1
12 tháng 8 2015
ta có 6n + 3 chia hết cho 3n +6
6n + 12 -9 ..................3n +6
2 .(3n + 6) -9 .................. 3n +6
9 ..................3n +6 ( vì 2. ( 3n +6 ) chia hết cho 3n +6)
Suy ra 3n + 6 thuộc tập hợp { -9, -3, -1, 1. 3. 9}
ta có bảng
| 3n + 6 | -9 | -3 | -1 | 1 | 3 | 9 |
| 3n | -15 | -9 | -7 | -5 | -3 | 3 |
| n | -5 | -6 | loại | loại | -1 | 1 |
NM
5
20 tháng 4 2017
Ta có:\(\frac{6N+3}{3N+6}=\frac{6N+12-9}{3N+6}=\frac{2\left(3N+6\right)-9}{3N+6}=2-\frac{9}{3N+6}\)
Để \(6N+3⋮3N+6.\)Thì \(9⋮3N+6\)
=>3N+6\(\in\)Ư(9)
=>3N+6\(\in\){1;3;9}
=>3N=3
=> N=3:3
=> N=1
Vậy N=1
DT
5 tháng 4 2019
Ta có: \(\frac{6n+3}{3n+6}=\frac{6n+12-9}{3n+6}=\frac{2\left(6n+3\right)-9}{3n+6}=2-\frac{9}{3n+6}\)
Để 6n+3 chia hết cho 3n+6. thì 9 chia hết cho 3n+6
=> 3n+6\(\in\)Ư(9)
=> 3n+6 \(\in\){1,3,9}
=> 3n = 3
=> n = 3:3
=> n = 1
TV
1
LS
17 tháng 12 2020
6n+3=6n+12-9=(6n+12)-9
để 6n+3 chia hết cho3n+6 thì
(6n+12)-9 chia hết cho3n+6
2(3n+6)-9 chia hết cho3n+6
vì 2(3n+6)chia hết cho3n+6
good luck!
nên- 9 phảichia hết cho3n+6
3n+6 thuộc ước của -9
3n+6 thuộc -1;-9;-3;1;3;9
5 tháng 7 2016
a,
(n+4)⋮n
Mà (n+4)=n+4
n⋮n
Suy ra còn lại 4 cũng phải chia hết cho n
=> 4⋮n
=> n∈U(4)={±1;±2;±4}
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