a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

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

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
 

\(2^{333}< 3^{222}\)