2\(^{91}\)>2\(^{90}\)=(2\(^5\))\(^{18}\)=32\(^{18}\)>25\(^{18}\)=(5\(^2\))\(^{18}\)=5\(^{36}\)>5\(^{35}\)

Vậy 2\(^{91}\)>5\(^{35}\)

ta có:
2^91 = (2^13)^7 = 8192^7
5^35 = (5^5)^7 = 3125^7
Vì 8192^7 > 3125^7 nên 2^91 > 5^35

Ta có: 2^91=(2^13)^7=8192^7

         5^35=(5^5)^7=3125^7

Vậy 2^91>5^35

Ta có: 
2^91 = (2^13)^7 = 8192^7 
5^35 = (5^5)^7 = 3125^7 
Vì 8192^7 > 3125^7 nên 2^91 > 5^35.

Học tốt ^-^

2^91 và 5^35

2^91=(2^13)^7=8192^7

5^35=(5^5)^7=3125^7

Vì 8192^7>>3125^7=> 2^91>5^35

Vậy 2^91>5^35

A, 11/19 < 13/18

B, 13/17 < 35/39

11/19<11/18<13/18 nen 11/19<13/18

\(2^{225}=8^{75}< 9^{75}=3^{150}\)

\(2^{91}>2^{90}=32^{18}>25^{18}=5^{36}>5^{35}\)

\(99^{20}=\left(99.99\right)^{10}< \left(99.101\right)^{10}=9999^{10}\)

a, \(2^{225}=\left(2^3\right)^{75}\) 

    \(3^{150}=\left(3^2\right)^{75}\)

b,\(2^{91}=\left(2^{13}\right)^7\)

\(5^{35}=\left(5^5\right)^7\)

c,\(99^{20}=\left(99\cdot99\right)^{10}\)

\(9999^{10}=\left(99\cdot101\right)^{10}\)

a, 2²²⁵ = ( 2³ )⁷⁵ = 8⁷⁵

3¹⁵⁰ = ( 3² )⁷⁵ = 9⁷⁵

Vì 8⁷⁵ < 9¹⁵

=> 2²²⁵ < 3¹⁵⁰

Vậy ...

b, 2⁹¹ > 2⁹⁰ 

Mà 2⁹⁰ = 32¹⁸ và 32¹⁸ > 25¹⁸

25 ¹⁸ = 5³⁶ và 5³⁶ > 5³⁵

=> 2⁹¹ > 2⁹⁰ > 25¹⁸ > 5³⁵

=> 2⁹¹ > 5³⁵

Vậy ...

c, 99²⁰ = ( 99² )¹⁰ = 9801¹⁰

Mà 9801 < 9999 

=> 9801¹⁰ < 9999¹⁰

=> 99²⁰ < 9999¹⁰

Vậy ...

Hok tốt

a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A=2+2^2+2^3+2^4+...+2^{201}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A-A=2^{101}-1\)

\(A=2^{201}-1\)

Ta có 2²⁰¹ > 2²⁰¹ - 1 => B > A => 2²⁰¹ > 1 + 2 + 2² + 2³ +...+ 1¹⁰⁰

b) 2¹⁰⁰ = 2³¹ . 2⁶³ . 2⁶ = 2³¹ . (2⁹)⁷ . (2²)³ = 2³¹ . 512⁷ . 4³ (1)

10³¹ = 2³¹ . 5²⁸ . 5³ = 2³¹ . (5⁴)⁷ . 5³ = 2³¹ . 625⁷ . 5³ (2)

Từ (1) , (2) => 2³¹ . 512⁷ . 4³ < 2³¹ . 625⁷ . 5³ ( vì 512⁷ < 625⁷ và 4³ < 5³ )

=> 2¹⁰⁰ < 10³¹

e) Ta có:

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰

10³⁰ = (10³)¹⁰ = 1000¹⁰
Vì 1024¹⁰ > 1000¹⁰ => 2¹⁰⁰ > 10³⁰