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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) 820 và 720
vì 8>7 nên 820>720
b) 420 và 1620
vì 4<16 nên 420<1620
c) 277= (33)7= 321
815=( 34)5=320
vì 21>20 nên 321>320 hay 277> 815
e) 521= 520 . 5
vì 520 . 5>520 . 4 nên 521> 4 . 520
Bài 1 :
a,820 > 720
b, 420 = 1610
c, 277 > 815
d , 554 > 381
e, 521 > 4 . 520
f, 220 > 7.217
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) \(7^8+7^9+7^{10}\)
\(=7^8\left(1+7+7^2\right)\)
\(=7^8.57⋮57\)
b) \(10^{10}-10^9-10^8=10^8\left(10^2-10-1\right)\)
\(=10^8⋮89\)
\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
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
nhìn thoy đã thấy nản r`....
a/ \(3^{500}=\left(3^5\right)^{100}=243^{100};5^{300}=\left(5^3\right)^{100}=125^{100}\)
Ta thấy \(243^{100}>125^{100}\Rightarrow3^{500}>5^{300}\)
b/ \(125^5=\left(5^3\right)^5=5^{15};25^7=\left(5^2\right)^7=5^{14}\)
ta thấy \(5^{15}>5^{14}\Rightarrow125^5>25^7\)
c/ \(9^{20}=\left(3^2\right)^{20}=3^{40};27^{13}=\left(3^3\right)^{13}=3^{39}\)
Ta thấy \(3^{40}>3^{39}\Rightarrow9^{20}>27^{13}\)
...còn lại tự lm nốt nhá....
a) 820 và 720
\(\Rightarrow\) 820 > 720
b) 420 và 1610
420
1610 = (42)10 = 420
\(\Rightarrow\) 420 = 1610
c) 277 và 815
277 = (33)7 = 321
815 = (34)5 = 320
\(\Rightarrow\) 277 > 815
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