bài 4 : c1 \(3^{4000}\)và \(9^{2000}\)

\(\Leftrightarrow9^{2000}\Leftrightarrow\left(3^2\right)^2^{000}\Leftrightarrow3^{4000}\)

vì \(3^{4000}=3^{4000}\Leftrightarrow3^{4000}=9^{2000}\)

c2 

ta có 

\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)

\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)

vì \(81^{1000}=81^{1000}\Leftrightarrow3^{4000}=9^{2000}\)

bài 5 

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

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

vì \(8^{111}< 9^{111}\Leftrightarrow2^{332}< 3^{223}\)

3) M = 2²⁰¹⁰ - (2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰)

Đặt N = 2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰

=> 2N = 2²⁰¹⁰ + 2²⁰⁰⁹ + .... + 2² + 2¹

=> 2N - N = (2²⁰¹⁰ + 2²⁰⁰⁹ + .... + 2² + 2¹) - (2²⁰⁰⁹ + 2²⁰⁰⁸ + ....  + 2¹ + 2⁰)

=> N = 2²⁰¹⁰ - 1

Khi đó M = 2²⁰¹⁰ - (2²⁰¹⁰ - 1) = 1

4) C1 Ta có 3⁴⁰⁰⁰ = (3⁴)¹⁰⁰⁰ = 81¹⁰⁰⁰ = (9²)¹⁰⁰⁰ = 9²⁰⁰⁰ 

3⁴⁰⁰⁰ = 9²⁰⁰⁰

C2 Ta có : 3⁴⁰⁰⁰ = (3⁴)¹⁰⁰⁰ = 81¹⁰⁰⁰ (1)

Lại có 9²⁰⁰⁰ = (9²)¹⁰⁰⁰ = 81¹⁰⁰⁰ (2)

Từ (1) (2) => 3⁴⁰⁰⁰ = 9²⁰⁰⁰

5 Ta có 2³³² < 2³³³ = (2³)¹¹¹ = 8¹¹¹ < 9¹¹¹ = (3²)¹¹¹ = 3²²² < 3²²³

=> 2³³² < 3²²³

2) Ta có n¹⁵⁰ < 5²²⁵

=> (n⁵)⁷⁵ < (5³)⁷⁵

=> n⁵ < 5³

=> n⁵ < 125

Vì n là số nguyên lớn nhất => n = 2

5.

Ta có:\(2^{332}\)<\(3^{223}\)

=\(\left(2^3\right)^{111}\)=\(8^{111}\)

=\(3^{223}\)>\(3^{222}\)

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

Ta có:8<9 ⇛:\(8^{111}< 9^{111}\)

vậy:\(2^{332}< 3^{223}\)

1.

M = 2²⁰¹⁰ - ( 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰ )

đặt N = 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰

2N = 2²⁰¹⁰ + 2²⁰⁰⁹ + ... + 2² + 2¹

2N - N = ( 2²⁰¹⁰ + 2²⁰⁰⁹ + ... + 2² + 2¹ ) - ( 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰ )

N = 2²⁰¹⁰ - 2⁰

Thay N vào ta được : 

M = 2²⁰¹⁰ - ( 2²⁰¹⁰ - 2⁰ )

M = 2²⁰¹⁰ - 2²⁰¹⁰ + 2⁰

M = 2⁰ = 1

2.

Ta có :

2³³² < 2³³³ = ( 2³ ) ¹¹¹ = 8¹¹¹

3²²³ > 3²²² = ( 3² ) ¹¹¹ = 9¹¹¹

Vì 2³³² < 8¹¹¹ < 9¹¹¹ < 3²²³

Ta có : 

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

Vì : 

\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)

\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)

Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)

                  \(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

    \(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)

   \(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)

nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)

hay A > B

Vậy A > B 

Bài 2 : 

\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2008}\)và 4