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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 => \(24^{10}+2.24^{10}< 27^{10}+64^{10}\Rightarrow3.24^{10}< 8^{10}+27^{10}+64^{10}\)
\(\Rightarrow3.24^{10}< 3^{30}+2^{30}+4^{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4^30=2^30*2^30
=2^30*4^15
3*24^10=3*3^10*8^10=3^11*2^30
mà 4^30>3^11
nên 2^30+3^30+4^30>3*24^10
Ta có: 4^30=2^30.2^30=2^30.4^15
3.24^10=3.(3.2^3)^10=2^30.3^11
Ta thấy: 3^11<3^15<4^15 => 4^15>3^11
Vì 4^15>3^11 nên 2^30.4^15>2^30.3^11
=>2^30+3^30+4^30>3.24^10
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2^24 > 3^16
b, 5^300>3 ^500
c,99^20 > 9999^10
d, 2^30 +3^44 +4^30 < 3x24^10
![](https://rs.olm.vn/images/avt/0.png?1311)
4^30=2^30*2^30
=2^30*4^15
3*24^10=3*3^10*8^10=3^11*2^30
mà 4^30>3^11
nên 2^30+3^30+4^30>3*24^10
![](https://rs.olm.vn/images/avt/0.png?1311)
4^30=2^30*2^30
=2^30*4^15
3*24^10=3*3^10*8^10=3^11*2^30
mà 4^30>3^11
nên 2^30+3^30+4^30>3*24^10
![](https://rs.olm.vn/images/avt/0.png?1311)
210 +320+420 =( 22)10 + (3^2)^10 + (4^2)^10 = 4^10 +9^10+16^10
210 +320+420 =( 22)10 + (3^2)^10 + (4^2)^10 = 4^10 +9^10+16^10
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
2) \(3^{21}=3^{20}\cdot3=9^{10}\cdot3\)
\(2^{31}=2^{30}\cdot2=8^{10}\cdot2\)
mà \(9^{10}\cdot3>8^{10}\cdot2\)=> tự viết tiếp
3) đợi chút
430 = (43)10 = 6410 > 4810 = ( 2 . 24 )10 = ( 210 ) . ( 2410 ) > 3 . 2410
=> 230 + 330 + 430 > 3 . 2410
.