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a/ \(3^{150}=\left(3^2\right)^{75}=9^{75}\)
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(9^{75}>8^{75}\Rightarrow3^{150}>2^{225}\)
b/
\(20162016^{10}=\left(2016.10001\right)^{10}=2016^{10}10001^{10}\)
\(2016^{20}=2016^{10}.2016^{10}\)
\(10001^{10}>2016^{10}\Rightarrow2016^{10}.10001^{10}>2016^{10}.2016^{10}\Rightarrow20162016^{10}>2016^{20}\)
c/ \(\frac{222^{333}}{333^{222}}=\frac{\left(222^3\right)^{111}}{\left(333^2\right)^{111}}=\frac{\left(2^3.111^3\right)^{111}}{\left(3^2.111^2\right)^{111}}=\left(\frac{8.111}{9}\right)^{111}\)
\(\frac{888}{9}>1\Rightarrow\left(\frac{888}{9}\right)^{111}>1\Rightarrow222^{333}>333^{222}\)
a) Ta có: 3^150 = 3^2.75 = (3^2)^75 = 9^75
2^225 = 2^3.75 = (2^3)^75 = 8^75
Vì 9 > 8 nên 9^75 > 8^75
Vậy 3^150 > 2^225
b) Ta có: 2016^20 = 2016^10+10 = 2016^10 . 2016^10
20162016^10 = (10001 . 2016)^10 = 10001^10 . 2016^10
Vì 2016^10 < 10001^10 nên 2016^10 . 2016^10 < 10001^10 . 2016^10
Vậy 2016^20 < 20162016^10
`@` `\text {Ans}`
`\downarrow`
Ta có:
\(5^{333}=\left(5^3\right)^{111}=125^{111}\)
\(11^{222}=\left(11^2\right)^{111}=121^{111}\)
Vì `125 > 121 =>`\(125^{111}>121^{111}\)
`=>`\(5^{333}>11^{222}\)
Vậy, \(5^{333}>11^{222}\)
_____
`@` So sánh lũy thừa cùng cơ số:
Nếu `m > n =>`\(a^m>a^n\left(m,n\ne0,a>1\right)\)
`@` So sánh lũy thừa cùng số mũ:
Nếu `a > b =>`\(a^m>b^m\left(a,b>1,m\ne0\right)\)
`@` `\text {Kaizuu lv uuu}`
\(333^{333}=3^{333}.111^{333}\)
\(555^{222}=5^{222}.111^2\)
\(3^{333}=27^{111}>5^{222}=25^{111}\) (1)
\(111^{333}>111^{222}\)(2)
Từ (1) và (2) \(\rightarrow333^{333}>555^{222}\)
ta có 2^333=(2^3)^111=8^111
3^222=(3^2)^111=9^111
vi 8^111<9^111
=>2^333<3^222
2^333 = (2^3)^111 = 8^111
3^222 = (3^2)^111 = 9 ^111
Vì 8< 9 => 8^111 < 9^111 => 2^333 < 3^222
Ta có: 2^333 = (2^3)^111 = 8^111
3^222 = (3^2)^111 = 9^111
Vì 8^111 < 9^111 nên 2^333 < 3^222
\(2^{333}=\left(2^3\right)^{111}=8^{111}\\ 3^{222}=\left(3^2\right)^{111}=9^{111}\)
VÌ\(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
a, \(2^{91}\) và \(5^{35}\)
Ta có :
\(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
Vì \(8192>3125\) nên \(2^{91}>5^{35}\)
b, \(222^{333}\) và \(333^{222}\)
Ta có :
\(222^{333}=\left(2.111\right)^{333}=2^{333}.111^{333}=\left(2^3\right)^{111}.111^{333}=8^{111}.111^{333}\)
\(333^{222}=\left(3.111\right)^{222}=3^{222}.111^{222}=\left(3^2\right)^{111}.111^{222}=9^{111}.111^{222}\)
Vì \(8^{111}< 9^{111}\) nên \(222^{333}< 333^{222}\)