(1+1/2)×(1+1/3)×(1+1/4)×....×(1+1/n)
Tim n biet n thuoc N va n lon hon hoac bang 2
cho A = 1+2+3+4+...+n va B = 2n + 1 ( voi n thuoc N , n lon hon hoac bang 2) chung minh A va B la hai so nguyen to cung nhau
Tïm n thuoc N sao cho
a) 24 : het 2n - 1(n lon hon hoac bang 1)
b) 4n - 7 : het 2n - 1 (n lon hon hoac bang 1)
So lien truoc cua so 3m -1 la so nao?( Voi m thuoc N va m lon hon hoac bang 2)
Số liền trước của 3m-1 là 3m-2 (Với m thuộc N và m lớn hơn hoặc băng 2 )
Tik cho mk nha bn ^^ May mắn cả năm
ai trl dc mik tick cho nha
1¦ viet tap A cac so N n biet :
52/13 < n < hoac = 56/7
2¦ tim n thuoc N sao cho cac phan so co gt la so Z
a. n+6 : n
b n : n-3
3¦ tim 2 phan so co hieu bang 7 biet 1/3 so nho = 1/4 so lon
nhanh nha mik dang can
CMR :\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{n^2}< 1\)
(n Thuoc N;n lon hon hoac = 2
Ta đặt:A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{n^2}\)
Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
....
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> A < \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{\left(n-1\right)n}\)
=> A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
=> A < \(1-\frac{1}{n}< 1\)(ĐPCM )
Vậy A < 1
tim n cac tich sau voi n thuoc N, n> hoac bang 2
(1-1/2)(1-1/30)(1-1/4)......(1-1/n)
(1+1/2)(1+1/3)(1+1/4)...(1+1/n)
tim x thuoc N biet
a, 125.n=5^7
b, 2^3.n-3^4=2^5-5
c, (2^3+2) n+3^2.n.5-10=10^2
d, 5^n.5=125
e, 9 be hon hoac bang 3^n<90
f, (3n+1)^3=64
tim x thuoc N biet
a, 125.n=5^7
b, 2^3.n-3^4=2^5-5
c, (2^3+2) n+3^2.n.5-10=10^2
d, 5^n.5=125
e, 9 be hon hoac bang 3^n<90
f, (3n+1)^3=64