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 bieta, 125.n 5 7b, 2 3.n 3 4 2 5 5c, 2 3 2 n 3 2.n.5 10 10 2d, 5 n=
125
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
a: =>n*5^3=5^7
=>n=5^4=625
c: \(\Leftrightarrow2\cdot3^n=3^4+2^5-5=81+32-5=108\)
=>3^n=54
=>\(n\in\varnothing\)
d: =>5^n=25
=>n=2
f: =>3n+1=4
=>3n=3
=>n=1
CHO S= 5+5^2+5^3+5^4+...+5^95+5^96
a.chung to s chia het cho 126
b. tim chu so tan cung cua s
11. tim nEn biet
a. 9<3^n be hon hoac bang 243
b. 9<3^n<27
c. 25 be hon hoac bang 5^n < 3125
tim n thuoc N
,25 be hun hoac bang 5n be hun hoac bang 125
Ta có
\(25\le5^n\le125\)=>\(5^2\le5^n\le5^3\)=>\(2\le n\le3\)=>n=2 hoặc n=3
5 mu n nhan 5 mu n cong 1 nhan 5 mu n cong 2 be hon hoac bang 10 mu 18
hay viet cac tap hop sau bang cach khac
A={ 9;11;13;...;99]
B={ 12 ; 24 ; 36 ;48}
M= { THANG 4 ; THANG 6 ;THANG 9 ; THANG 11}
N = { X THUOC N / X LON HON HOAC BANG 3 , LON HON HOAC BANG 10}
Q = { X THUOC N / X LON HON 0 ,BE HON 10}
(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
tim x thuoc N : 8^x<512^4 2^x+1.2^x+2.2^x+3 be hon hoac bang 10...0 : 5^24 ( 10...0 co 24 chu so 0 )
cac ban giai nhanh gium mik nha , ti nua mik phai nop roi . cam on
8^x=2^3.x
512=2^9
512^4=2^36
=> 3.x<36
x<9
x=(0,1,2,3,4,5,6,7,8)