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\(2^{50}=\left(2^5\right)^{10}=32^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Suy ra: 250 > 520
b)
\(9^{200}=\left(9^2\right)^{100}=81^{100}\)
Suy ra: 99100 > 81100
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\(99^{100}=99^{100}\)
\(33^{200}=\left(33^2\right)^{100}=1089^{100}\)
vi \(1089>99\) nen \(1089^{100}>99^{100}\)
vay \(99^{100}< 33^{200}\)
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Viết rối qá chả thấy j.
\(99^2vs9999^{10}\)
\(9999^{10}=\left(101\cdot99\right)^{10}=101^{10}\cdot99^{10}\)
Vì \(99^{10}>99^2=>99^2< 9999^{10}\)
a) Ta có: 2^91 = (2^13)^7 = 8192^7
5^35 = (5^5)^7 = 3125^7
Vì 8192 > 3125 nên 8192^7 > 3125^7
Vậy 2^91 > 5^35
b) Ta có: 9999^10 = 99^10 . 101^10
Vì 99^2 < 99^10 nên 99^2 < 99^10 . 101^10
Vậy 99^2 < 9999^10
c) Ta có: 2^300 = (2^6)^50 = 64^50
3^200 = (3^4)^50 = 81^50
Vì 49 < 64 < 81 nên 49^50 < 64^50 < 81^50
Vậy 49^50 < 2^300 < 3^200
d) 9^3/25^3 = (9/25)^3
3^6/2^12 = (3^2)^3/(2^4)^3 = 9^3/16^3 = (9/16)^3
Vì 9/25 < 9/16 nên (9/25)^3 < (9/16)^3
Vậy 9^3/25^3 < 3^6/2^12.
ta có : 9^200=(9^2)^100=81^100.
vì 81^100<99^100 nên 9^200<99^100.
\(9^{200}=\left(9^2\right)^{100}=81^{100}< 99^{100}\)
Vậy \(9^{200}< 99^{100}\)