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DD
Đoàn Đức Hà
Giáo viên
8 tháng 12 2021
a) \(\left(n+1\right)^2+\left(n+2\right)^2+\left(n+3\right)^2=\left(n+10\right)^2\)
\(\Leftrightarrow n^2+2n+1+n^2+4n+4+n^2+6n+9=n^2+20n+100\)
\(\Leftrightarrow2n^2-8n-86=0\)
\(\Leftrightarrow n^2-4n=43\)
Ta có: \(n^2-4n=n^2-n-3n=n\left(n-1\right)-3n\)
\(n\left(n-1\right)\)là tích hai số tự nhiên liên tiếp nên khi chia cho \(3\)dư \(0\)hoặc \(2\).
Suy ra \(n^2-4n\)chia cho \(3\)dư \(0\)hoặc \(2\).
Mà \(43\)chia cho \(3\)dư \(1\)
do đó phương trình đã cho không có nghiệm tự nhiên.
b) Ta có: \(n^2+h^2+b^2+k^2+n+h+b+k=\left(n^2+n\right)+\left(h^2+h\right)+\left(b^2+b\right)+\left(k^2+k\right)\)
\(=n\left(n+1\right)+h\left(h+1\right)+b\left(b+1\right)+k\left(k+1\right)\)chia hết cho \(2\).
mà \(n+h+b+k\)chia hết cho \(6\)nên chia hết cho \(2\)
suy ra \(n^2+h^2+b^2+k^2\)chia hết cho \(2\)suy ra không phải là số nguyên tố
(do \(n^2+h^2+b^2+k^2>2\)).
NH
3
12 tháng 8 2016
Với n lẻ thì an+bn=(a+b)( an-1-an-2.b+an-3.b2-...-a.bn-2+bn-1) hay với n lẻ thì an+bn chia hết cho a+b
1n+2n+3n+4n=(1n+4n)+(2n+3n)
Áp dụng phần trên thì với n lẻ (1n+4n) chia hết cho 5 , 2n+3n chia hết cho 5
Kết luận : n lẻ
B
1
MN
3
11 tháng 11 2015
n2+3 chia hết cho n+2
n2+2n-2n+3
n(n+2)-(2n-3) chia hết cho n+2
=>2n+4-7 chia hết cho n+2
2(n+2)-7 chia hết cho n+2
=>7 chia hết cho n+2 hay n+2 EƯ(7)={7}
=>n=5
Vậy để n2+3 chia hết cho n+2 thì n=5
LD
25 tháng 2 2017
ta có:n2+3=n.(n+2)-2.(n+2)+7 chia hết cho n+2
mà n.(n+2)-2.(n+2)chia hết cho n+2
suy ra 7 chia hết cho n+2
suy ra n+2 thuộc tập hợp các ước của 7=7
suy ra n=5
vậy n=5
TB
1
NN
1
ko bt và ko CARE
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