2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

A=1+"2+2²+2³+...+2^2014"

2A="2+2²+2³+...2²⁰¹⁴"+2²⁰¹⁵

=>A=2A-A=2²⁰¹⁵-1(do 2 phần có dấu(")trừ cho nhau là hết nên còn 2²⁰¹⁵-1)

Vì 2²⁰¹⁵-1=2²⁰¹⁵-1

nên A=B

! ) A = (3999 /2 +1 ) + ( 3998/ 3 + 1 ) + ( 3997 / 4 + 1 ) +...+ ( 1/ 4000 + 1 ) + 1

 (Ta lấy 4000/1 = 4000  rải đều  1, 1 ,1 cho 3999 phân số  và dư lại 1  = 4001/4001 )

= 4001 /2 + 4001 / 3 + 4001 /4 + ...+ 4001 /4000 + 4001 / 4001

= 4001 ( 1/2 + 1/3 + 1/4 +..+ 1/ 4001 )  vay A: B = 4001

a,A=2^0+2^1+2^2+...+2^2014

2A=2^1+2^2+2^3+...+2^2015

2A-A=(2^1+2^2+2^3+...+2^2015)-(2^0+2^1+2^2+...+2^2014)

A=2^2015-2^0=2^2015-1=B

=>A=B

b,A=2014.2016=2014.(2015+1)=2014.2015+2014

B=2015^2=2015.2015=(2014+1).2015=2014.2015+2015

Vì 2014<2015 => A<B.

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)