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Bài 1:
\(\left(x-2013\right)^{2014}=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2013=1\\x-2013=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2014\\x=2012\end{cases}}}\)
Vậy x=2014; x=2012
Bài 2:
a) Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Ta thấy 8<9 => \(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
b) Ta có: \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Ta thấy \(3^{2009}< 3^{2010}\Rightarrow3^{2009}< 9^{1005}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Thấy \(9801< 9999\Rightarrow9801^{10}< 9999^{10}\Rightarrow99^2< 9999^{10}\)
B1: (x-2013)2014=1 =>x-2013=1;-1=>x=2014;2012 B2: a)có:2333=(23)111=8111 ; 3222=(32)111=9111 =>2333<3222(8111<9111) b)có:91005=(32)1005=32010 >32009 =>91005>32009 c)có:9920=(992)10=980110<999910 =>9920<999910
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) \(9^{1005}=\left(3^2\right)^{1005}=3^{2.1005}=3^{2010}>3^{2009}\)
\(A,2^{333}\) và \(3^{222}\)
Ta có:
\(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì 8<9 \(\Rightarrow8^{111}< 9^{111}\)
\(\Rightarrow2^{333}< 3^{222}\)
B,\(3^{2009}\) và \(9^{2005}\)
Ta có:
\(9^{2005}=\left(3^2\right)^{2005}=3^{4010}\)
Vì 2009 < 4010 \(\Rightarrow3^{2009}< 3^{4010}\)
\(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
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: \(8100^{10}< 9999^{10}\)
\(\Rightarrow90^{20}< 9999^{10}\)
\(2^{225}=8^{75}< 9^{75}=3^{150}\)
\(2^{91}>2^{90}=32^{18}>25^{18}=5^{36}>5^{35}\)
\(99^{20}=\left(99.99\right)^{10}< \left(99.101\right)^{10}=9999^{10}\)
a, \(2^{225}=\left(2^3\right)^{75}\)
\(3^{150}=\left(3^2\right)^{75}\)
b,\(2^{91}=\left(2^{13}\right)^7\)
\(5^{35}=\left(5^5\right)^7\)
c,\(99^{20}=\left(99\cdot99\right)^{10}\)
\(9999^{10}=\left(99\cdot101\right)^{10}\)
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 : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)
Do : \(3^{2009}< 3^{2010}\left(2009< 2010\right)\)
\(\Rightarrow3^{2009}< 9^{1005}\)