a, Ta có:

\(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)

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

b, Ta có:

\(2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\)nên \(2^{91}>5^{35}\)

2^225>3^150

2^91<5^35

99^20<9999^10

a, 2²²⁵ và 3¹⁵⁰
2²²⁵ = ( 2³ )⁷⁵ = 8⁷⁵
3¹⁵⁰ = ( 3² )⁷⁵ = 9⁷⁵
Do 8⁷⁵ < 9⁷⁵ nên 2²²⁵ < 3¹⁵⁰ 
b,c tương tự
 

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

\(d,10^{20}=\left(10^2\right)^{10}=100^{10}\\ Vì:100^{10}>20^{10}\left(Do:100>20\right)\\ \Rightarrow10^{20}>20^{10}\)

a) Ta có: \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)

                \(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)

\(\Rightarrow8^{75}< 9^{75}\)\(\Rightarrow2^{225}< 3^{150}\)

b) Ta có : \(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\)

                \(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)

\(\Rightarrow8192^7>3125^7\)\(\Rightarrow2^{91}>3^{35}\)

c) Ta có: \(99^{20}=99^{2.10}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)

               \(9999^{10}=\left(99.101\right)^{10}\)

Vì 99<101 \(\Rightarrow\left(99.99\right)^{10}< \left(99.101\right)^{10}\)\(\Rightarrow99^{20}< 9999^{10}\)

Ta có 2²²⁵ = (2³)⁷⁵ = 8⁷⁵

3¹⁵⁰ = (3²)⁷⁵ = 9⁷⁵

Vậy 2²²⁵ < 3¹⁵⁰

Ta có:

2⁹¹ = (2¹³)⁷ = 8192⁷

5³⁵ = (2⁵)⁷ = 3125⁷ 

Vì 8192⁷ > 3125⁷

=> 2⁹¹ > 5³⁵

Ta có:

99²⁰ = 99¹⁰.99¹⁰ < 99¹⁰.100¹⁰ < 99¹⁰.101¹⁰ = 9999¹⁰

=> 99²⁰ < 9999¹⁰

2²⁵⁵ > 3¹⁵⁰

2⁹¹ < 5³⁵

99²⁰ < 9999¹⁰

bạn dùng máy tính casio ccũng được

Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)

Vì \(9801< 9999\)nên \(9801^{10}=9999^{10}\)

Vậy \(99^{20}< 9999^{10}\)

a, 2^27 = 2^3.9 = 8^9 

3^18 = 3^2.9 = 9^9 

vì 8<9 => 8^9 < 9^9 => 2^27 < 3^18

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

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

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

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 ...

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