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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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HB
0
HO
0
NM
0
21 tháng 9 2018
a/ \(\frac{1}{n\left(n-1\right)\left(n+1\right)}=\frac{1}{n^3-n}>\frac{1}{n^3}\)
b/ \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n^3+3n^2+2n}< \frac{1}{n^3}\)
c/ Ap dụng câu b ta được
\(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2006^3}>\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2006.2007.2008}\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2006.2007}-\frac{1}{2007.2008}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{2007.2008}\right)>\frac{1}{12}>\frac{1}{15}\)
DX
2
25 tháng 6 2017
A\(=n^4-4n^3-4n^2+16n\)
\(=\left(n^4-4n^2\right)+\left(-4n^3+16n\right)\)
\(=n^2\left(n^2-4\right)-4n\left(n^2-4\right)\)
\(=n\left[\left(n^2-4\right)\left(n-4\right)\right]\)
\(n.\left(n+2\right)\left(n-2\right)\left(n-4\right)\)
Ta có: tích 4 số chắn liên tiếp chia hết cho 384
=> đpcm
NT
25 tháng 6 2017
n chẵn => n=2k
\(\Rightarrow A=\left(2k\right)^4-4.\left(2k\right)^3-4\left(2k\right)^2+16.2k\\ =16k^4-32k^3-16k^2+32k\\ =16k^3\left(k-2\right)-16k\left(k-2\right)\\ =\left(k-2\right)\left(16k^3-16k\right)\\ =\left(k-2\right)\left(16k\left(k^2-1\right)\right)\\ =16.\left(k-2\right)\left(k-1\right).k.\left(k+1\right)\\ \)
Tích 4 số tự nhiên liên tiếp luôn chia hết cho 3;8 nên chia hết cho 24
\(\Rightarrow A⋮16.24\\ \Rightarrow A⋮384\)
CH
Cô Hoàng Huyền
Admin
VIP
4 tháng 4 2018
1. Nếu AB = AC:
Xét tam giác ABN và tam giác ACM có:
AN = AM (gt)
AB = AC (gt)
Góc A chung
\(\Rightarrow\Delta ABN=\Delta ACM\left(c-g-c\right)\)
\(\Rightarrow BN=CM\) (Hai cạnh tương ứng)
2.
a) Trên cạnh AB lấy điểm M' sao cho AM' = AC.
Ta có ngay \(\Delta AM'N=\Delta ACM\left(c-g-c\right)\)
\(\Rightarrow MC=NM'\)
Lại có AM' < AB nên NM' < NB
Vậy nên BN > CM
b) Ta thấy ngay MK > KN mà BN > MC nên BK = BN - KN > KC = MC - MK
2 tháng 7 2022
a: Xet ΔADB và ΔADE có
AB=AE
góc BAD=góc EAD
AD chung
DO đó: ΔADB=ΔADE
b: XétΔABC có AD là phân giác
nên DB/AB=DC/AC
mà AB<AC
nên DB<DC
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