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

99²⁰=99²⁰

9999¹⁰=(99¹⁰¹)¹⁰=99¹¹¹

99²⁰<99¹¹¹

Vậy 9²⁰<9999¹⁰

ta có 9999= 99 .101. 
do đó \(9999^{10}\) = \(99^{10}\) * \(101^{10}\)
còn \(99^{20}\) = \(99^{10}\) * \(99^{10}\)
vì \(99^{10}\) < \(101^{10}\) nên \(99^{10}\) * \(99^{10}\) < \(99^{10}\) * \(101^{10}\). 
vậy \(99^{20}\) < \(9999^{10}\). 
 

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

\(9801< 9999=>9801^{10}< 9999^{10}\)

hay \(99^{20}< 9999^{10}\)

Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

 ta có 9999= 99 *101. 
do đó 9999^10 = 99 ^10 * 101^10 
còn 99^20 = 99^10 * 99^10 
vì 99^10 < 101^10 nên 99^10 * 99^10 < 99 ^10 * 101^10 . 
vậy 99^20 < 9999^10. 
chào bạn

cảm ơn

a,

15^12=(3*5)^12=3^12*5^12

81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15

Vì 12<15 suy ra 5^12<5^15

Suy ra 3^12*5^12<3^12*5^15

\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)

\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)

\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)

Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)

\(2^{50}=\left(2^5\right)^{10}=32^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}\)

Suy ra: 2⁵⁰ > 5²⁰

b)

\(9^{200}=\left(9^2\right)^{100}=81^{100}\)

Suy ra: 99¹⁰⁰ > 81¹⁰⁰

\(5^{202}=\left(5^2\right)^{101}=25^{101}\)

\(2^{505}=\left(2^5\right)^{101}=32^{101}\)

Suy ra: 5²⁰² < 2⁵⁰⁵