Ta có: \(5^{x}\cdot5^{x+1}\cdot5^{x+2}=10\ldots0:2^{18}\) (18 chữ số 0)

=>\(5^{x+x+1+x+2}=10^{18}:2^{18}=5^{18}\)

=>\(5^{3x+3}=5^{18}\)

=>3x+3=18

=>3x=15

=>x=5

a: =>2^4x<2^28

=>4x<28

=>x<7

b: =>5^3x+3<5

=>3x+3<1

=>3x<-2

=>x<-2/3

a) \(16^x< 128^4\)

= (24)^x < (27)⁴

= 2^4x < 2²⁸

= 4x < 28

= x < 7 

Vậy \(x=\left\{0;1;2;3;4;5;6;\right\}\)

\(#Tuyết\)

\(2^x\cdot2^{x+1}\cdot2^{x+2}=100...00:5^{18}\)

\(2^{3x+3}=10^{18}:5^{18}\)

\(2^{3x+3}=2^{18}\)

\(\Rightarrow3x+3=18\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

a)\(\left(2n-1\right)^3=5^3\)

\(\Leftrightarrow2n-1=5\Rightarrow n=3\)

b)\(n^{20}=1\)(vì \(1^n=1\))

\(\Rightarrow n=1\)

c)\(5^{3n+3}\le\dfrac{10^{18}}{2^{18}}\)

\(\Leftrightarrow5^{3n+3}\le5^{18}\)

\(\Leftrightarrow3n+3\le18\)

\(\Rightarrow n\le5\)

10000...0+8=1000...08 (có 2014 chữ số 0)

\(1000...08⋮2\)

\(1000...08⋮9\)

2 và 9 là 2 số nguyên tố cùng nhau nên \(1000...08⋮18\)

a: \(\left(5x-10\right):\left(100x^2+18\right)=0\)

=>5x-10=0

=>5x=10

=>x=2

b: (-300):20+5(3x-1)=25

=>-15+5(3x-1)=25

=>5(3x-1)=25+15=40

=>3x-1=8

=>3x=9

=>x=3

c: \(\left(2021x+12\right)^2\left(x+4\right)>0\)

mà \(\left(2021x+12\right)^2>0\forall x\) nguyên

nên x+4>0

=>x>-4

mà x nguyên

nên x∈{-3;-2;-1;...}