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a)\(\left(x-2013\right)^{2014}=1\Leftrightarrow\left[{}\begin{matrix}x-2013=1\\x-2013=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2014\\x=2012\end{matrix}\right.\)
b) \(2^{333}=\left(2^3\right)^{111}=8^{111}< 9^{111}=\left(3^2\right)^{111}=3^{222}\)
\(3^{2009}< 3^{2010}=\left(3^2\right)^{1005}=9^{1005}\)
\(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Có: \(8^{111}< 9^{111}\)
\(\Leftrightarrow2^{333}< 3^{222}\)
\(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Có: \(3^{2010}>3^{2009}\)
\(\Rightarrow9^{1005}>3^{2009}\)
\(90^{20}=\left(90^2\right)^{10}=8100^{10}\)
Có: \(8100^{10}< 9999^{10}\)
\(\Rightarrow90^{20}< 9999^{10}\)
a, Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
b, Ta có : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
\(\Rightarrow3^{2009}< 9^{1005}\)
c, Ta có : \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
a) Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì 9>8 nên 9111>8111
Vậy 3222>2333
b) Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Vì 2010>2009 nên 32010>32009
Vậy 91005>32009
c)Ta có:\(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Vì 99<101 nên (99.99)10<(99.101)10
Vậy 9920<999910
a) \(2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8< 9\)\(\Rightarrow8^{111}< 9^{111}\)\(\Rightarrow2^{333}< 3^{222}\)
b) \(9^{1005}=\left(3^2\right)^{1005}=3^{2.1005}=3^{2010}>3^{2009}\)
Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Do : \(8^{111}< 9^{111}\left(8< 9\right)\)
\(\Rightarrow2^{333}< 3^{222}\)
ta có
9^1005=(3^2)^1005
=3^2010
vi 3^2010>3^2009
=>3^2009<9^1005
Ta có :
9^1005 = ( 3^2 ) ^ 1005 = 3^2010
Vì 3^2009 < 3^2010
=> 3^2009 < 9^1005
a: \(2^{333}=8^{111}< 9^{111}=3^{222}\)