Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a 5.125.625=5.5^3.5^4=5^8
b 10.100.1000=10.10^2.10^3=10^6
c 8^4.16^5.32=2^3^4.2^4^5.2^5=2^12.2^20.2^5=2^37
a) = \(5^1\cdot5^3\cdot5^4=5^{1+3+4}=5^8\)
b) = \(10^1\cdot10^2\cdot10^3=10^{1+2+3}=10^6\)
c) = \(2^{12}\cdot2^{20}\cdot2^5=2^{12+20+5}=2^{37}\)
\(10^{30}< 2^{100}\)
\(125^5>25^7\)
\(9^{20}< 27^{13}\)
\(3^{54}< 2^{81}\)
\(5^{40}< 620^{10}\)
\(3^{484}< 4^{636}\)
a) Có \(3^{125}=3^{124}.3=\left(3^4\right)^{31}.3=81^{31}.3\)
\(4^{93}=\left(4^3\right)^{31}=64^{31}\)
Vì \(81^{31}>64^{31}\Rightarrow81^{31}.3>64^{31}\)
=) \(3^{125}>4^{93}\)
b) Có \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Vì \(\frac{-7}{10^{2005}}=\frac{-7}{10^{2005}},\frac{-7}{10^{2006}}=\frac{-7}{10^{2006}},\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
=) \(\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}>\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
=) A > B
a) Ta có: 3124= (34)31= 8131
493= (43)31= 64 31
Do 8131 > 64 31 => 3124 < 493
Mà 3124< 3125 => 3125 > 493
a) Có: \(3+3^2+3^3+3^4+...+3^{99}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\\ =\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+...+3^{97}\left(3+3^2+3^3\right)\\ =39+3^3\cdot39+...+3^{97}\cdot39\\ =13\cdot3+3^3\cdot13\cdot3+...+3^{97}\cdot13\cdot3\\ =13\left(3+3^4+...+3^{98}\right)⋮13\left(đpcm\right)\)
b) Có: \(81^7-27^9-9^{13}\\ =\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\\ =3^{28}-3^{27}-3^{26}\\ =3^{26}\left(3^2-3-1\right)\\ =3^{24}\cdot\left(3^2\cdot5\right)\\ =3^{24}\cdot45⋮45\left(đpcm\right)\)
c) Có: \(24^{54}\cdot54^{24}\cdot2^{10}\\ =\left(2^3\cdot3\right)^{54}\cdot\left(2\cdot3^3\right)^{24}\cdot2^{10}\\ =2^{162}\cdot3^{54}\cdot2^{24}\cdot3^{72}\cdot2^{10}\\ =2^{196}\cdot3^{126}\\ =2^7\cdot\left(2^{189}\cdot3^{126}\right)\\ =2^7\cdot\left[\left(2^3\right)^{63}\cdot\left(3^2\right)^{63}\right]\\ =2^7\left(8^{63}\cdot9^{63}\right)\\ =2^7\cdot72^{63}⋮72^{63}\left(đpcm\right)\)
a) ta có: 3 + 32 + 33 + 34 + ... + 399
= (3 + 32 + 33) + (34 + 35 +36) + ... + (397 + 398 + 399)
= 3(1 + 3 + 32) + 34(1 + 3 + 3) + ... + 396(1 + 3 + 3)
= 3.13 + 34.13 + ... + 396.13
= 13(3 + 34 + ... + 396) ⋮ 13
vậy (3 + 32 + 33 + 34 + ... + 399) ⋮ 13
b) ta có: 817 - 279 - 913
= (34)7 - (33)9 - (32)13
= 328 - 327 - 326
= 326(32 - 3 - 1)
= 326 . 5 = 324 (9.5) = 324 . 45 ⋮ 45
Vậy (817 - 279 - 913) ⋮ 45
c) ta có: 2454.5424.210
= (23.3)54 . (2.33)24 . 210
= 2162 . 354 . 224 . 372 . 210
= 2196 . 3126
= (2193.3124).(23.32)
= (2193.3124).72 ⋮ 72
vậy (2454.5424.210) ⋮ 72
a) \(7^8+7^9+7^{10}\)
\(=7^8\left(1+7+7^2\right)\)
\(=7^8.57⋮57\)
b) \(10^{10}-10^9-10^8=10^8\left(10^2-10-1\right)\)
\(=10^8⋮89\)
a) \(125^5=\left(5^3\right)^5=5^{3\cdot5}=5^{15}\)
\(25^7=\left(5^2\right)^7=5^{2\cdot7}=5^{14}\)
\(5^{15}>5^{14}\Rightarrow125^5>25^7\)
b) \(3^{54}=\left(3^2\right)^{27}\)
\(2^{81}=\left(2^3\right)^{27}\)
\(3^2>2^3\Rightarrow3^{54}>2^{81}\)
d)
5^40 = ( 5^4)^10 = 625^10
mà 625^10 > 620^10 => 5^40 > 620^10
vậy ............
c)
10^30 = (10^3)^10 = 1000^10
2^100 = (2^10)^10 = 1024^10
mà 1000^10 < 1024^10 => 10^30 < 2^100
k mik nha!
a)
354=(36)9=7299
281=(39)9=196839
Vì 196839>7299
=>354<281
còn lại tự làm
a) Ta có: \(125^5=\left(5^3\right)^5=5^{15}\)
\(25^7=\left(5^2\right)^7=5^{14}\)
Ta thấy: 15 > 14 => 515 > 514
Vậy 1255 > 257
b) \(9^{20}=\left(3^2\right)^{20}=3^{60}\)
\(27^{13}=\left(3^3\right)^{13}=3^{39}\)
Vì 60 > 39 => 360 > 339
Vậy 920 > 2713
c) \(3^{54}=3^{2.27}=3^2.3^{27}=9.3^{27}\)
\(2^{81}=2^{3.27}=2^3.2^{27}=8.2^{27}\)
Vì 9 > 8 và 327 > 227
Vậy 354 > 281