CM:(n!)^2>n^n
K
Khách
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Những câu hỏi liên quan
9 tháng 8 2020
Câu 1:
Ta có: \(55^{n+1}+55^n\)
\(=55^n\left(55+1\right)=55^n\cdot56⋮56\)(đpcm)
Câu 2:
Ta có: \(5^6-10^4=\left(5^3-10^2\right)\left(5^3+10^2\right)\)
\(=\left(5^2\cdot5-5^2\cdot2^2\right)\cdot\left(5^2\cdot5+5^2\cdot2^2\right)\)
\(=5^2\cdot\left(5-2^2\right)\cdot5^2\cdot\left(5+2^2\right)\)
\(=5^4\cdot9=5^3\cdot45⋮45\)(đpcm)
DV
0
25 tháng 7 2017
Ta có \(\left(2-n\right)\left(n^2-3n+1\right)+n\left(n^2+12\right)+8\)
\(=2n^2-6n+2-n^3+3n^2-n+n^3+12n+8\)
\(=5n^2+5n+10=5\left(n^2+n+2\right)⋮5\)với mọi n
Vậy \(\left(2-n\right)\left(n^2-3n+1\right)+n\left(n^2+12\right)+8\)chia hết cho 5
S
3
22 tháng 12 2016
1. Ta có: n2 + n + 1 = n(n+1) + 1
Do n(n+1) chẵn với \(\forall n\in N\)=> n(n+1) + 1 lẻ => n2 + n + 1 không chia hết cho 4
Lại có : n(n+1) có tận cùng là 0 ; 2 ; 6 => n(n+1) + 1 có tận cùng là 1 ; 3 ; 7 => n2 + n + 1 không chia hết cho 5
2. mình không hiểu đề ^_^
MN
2
TC
10 tháng 7 2016
a) cách 1
2^4n = (24)n = ......6( có chữ số tận cùng là 6
=> (2^4n+1)+3= ......0( có chữ số tận cùng là 0)
=>(2^4n+1)+3 chia hết cho 5 với mọi n thuộc N?
cách 2
(2^4n+1)+3
=2*(24)n+3
=2*16n+3
=2(15 + 1)n+3
=2(5K+1) +3(với K là một số tự nhiên thuộc N)
=10K+5 chia hết cho 5
b ) áp dụng vào giống bài a thay đổi số thôi là đc
k mk nha!!!^~^
10 tháng 7 2016
Ta có : (24.n+1)+3 = (.....6) + 1) + 3 = (.....0)
=> (24.n+1)+3 có chữ số tận cùng là 0
=> (24.n+1)+3 chia hết cho 5
31 tháng 1 2016
Ta có:
\(\left(n+1\right)\left(n-1\right)=\left(n+1\right)n-\left(n+1\right)=n^2+n-n-1=n^2-1\)
\(\Rightarrow n^2\left(n+1\right)\left(n-1\right)=n^2\left(n^2-1\right)=n^4-n^2\)
Vậy \(n^4-n^2=n^2\left(n+1\right)\left(n-1\right)\)
HP
31 tháng 1 2016
Bài mk cũng gần giống Dung:
VP=(n+1)(n-1)=n.(n-1)+1(n-1)=n^2-n+n-1=n^2+0-1=n^2-1
Khi đó n^2.(n+1(n-1)=n^2.(n^2-1)=n^4-n^2=VT
=>VT=VP (đpcm)
TA
31 tháng 8 2020
Xét số nguyên dương thỏa mãn điều kiện \(1\le k< n-1\)
\(\Leftrightarrow n-k-1>0\Leftrightarrow nk-k^2-k>0\Leftrightarrow nk-k^2+n-k-n>0\)
\(\Leftrightarrow k\left(n-k\right)+n-k>n\Leftrightarrow\left(k+1\right)\left(n-k\right)>n\)
Lần lượt cho k = 1, 2, 3, ..., ( n - 2 ):
Với n > 2, ta có: \(2\left(n-1\right)>n\)
\(3\left(n-2\right)>n\)
\(4\left(n-3\right)>n\)
\(................\)
\(\left(n-1\right)\left[n-\left(n-2\right)\right]>n\)
\(\Leftrightarrow2.3.4...\left(n-1\right).2.3.4...\left(n-1\right)>n^{n-2}\)
\(\Leftrightarrow\left[2.3.4...\left(n-1\right)\right]^2>n^{n-2}\)
\(\Leftrightarrow\left[\left(n-1\right)!\right]^2>n^{n-2}\)
Nhân 2 vế với \(n^2\), ta có: \(\left(n!\right)^2>n^2\left(đpcm\right)\)
DA
31 tháng 8 2020
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