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a) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b) Ta có: \(2^{31}=\left(2\frac{31}{21}\right)^{21}=2,7822^{21}< 3^{21}\Rightarrow2^{31}< 3^{21}\)
c) Ta có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(4^{30}=\left(4^3\right)^{10}=64^{10}\)
Lại có: \(3.24^{10}=2.24^{10}+24^{10}\Rightarrow24^{10}< 27^{10}\left(1\right)\)
\(2.24^{10}< 48^{10}< 64^{10}\left(2\right)\)
Từ 1,2 => \(24^{10}+2.24^{10}< 27^{10}+64^{10}\Rightarrow3.24^{10}< 8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3.24^{10}< 3^{30}+2^{30}+4^{30}\)
\(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
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: \(8100^{10}< 9999^{10}\)
\(\Rightarrow90^{20}< 9999^{10}\)
2333=(23)111=8111
3222=(32)111=9111
Vì: 8<9 nên: 8111<9111
vậy: 2333<3222
b, 9920=(992)10=980110
Mà: 9801<9999 nên:
9920<999910
aTa có:
2333=(23)111=8111
3222=(32)111=9111
Do 8111<9111
=>2333<3222
b,Ta có:
9920=(992)10=980110
Do 980110 <999910
=>9920<999910
Xin lỗi bạn mình chỉ làm đc câu 2 thuiiii :((((((
b) Ta có:
\(555^{20}=111^{20}.5^{20}=111^{20}.\left(5^2\right)^{10}=111^{20}.25^{10}\)
\(222^{50}=111^{50}.2^{50}=111^{50}.\left(2^5\right)^{10}=111^{50}.32^{10}\)
Vì \(111^{50}.32^{10}>111^{20}.25^{10}\)nên \(222^{50}>555^{20}\)
\(2^{20}+3^{30}+4^{30}=4^{10}+9^{10}+64^{10}<64^{10}+64^{10}+64^{10}=3.64^{10}\)
\(324^{10}>320^{10}=\left(5.64\right)^{10}=5^{10}.64^{10}>3.64^{10}\)
\(\Rightarrow2^{20}+3^{30}+4^{30}<324^{10}\)
a, 920=(92)10=8110
vì 81 <9999 suy ra 920<999910
b, vì 3>2 suy ra 321>221