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a/ \(27^{11}=\left(3^3\right)^{11}=3^{33}\); \(81^8=\left(3^4\right)^8=3^{32}< 3^{33}\Rightarrow81^8< 27^{11}\)
b/ \(3^{2n}=\left(3^2\right)^n=9^n\); \(2^{3n}=\left(2^3\right)^n=8^n< 9^n\Rightarrow2^{3n}< 3^{2n}\)
a. 2711= (33)11 = 333
818 = (34)8 = 332
Suy ra 333>332 hay 2711>818
b. 32n = (32)n = 9n
23n = (23)n = 8n
Mà 9>8 suy ra 9n>8n hay 32n>23n
c. 523 = 522 . 5
(6.5)22 = 622 . 522
Vì 622>5 suy ra 522 . 5<622 . 522 hay 523<(6.5)22
d. 7245-7244 = 7244(72-1) = 7244 . 71
7244-7243 = 7243(72-1) = 7243 . 71
Vì 7244>7243 suy ra 7244 . 71>7243 . 71 hay 7245-7244>7244-7243

a) Ta có:
16^19=(24)19=276 ; 825=(23)25=275
Vì 76>75 nên 276>275. Vậy 1619>825
b) Ta có:
7245-7244=72(7244-7243)
Vậy 7245-7244 > 7244-7243
c) chịu
a, Ta có:16^19=(2^4)^19=2^76
8^25=(2^3)^25=2^75
Vì 2^75<2^76 nên 8^75<16^19
b,Ta có:72^45-72^74=72(72^44-72^73)
=>72^45-72^44>72^44-72^43
c,MÌNH GIẢI PHẦN NÀY SAU NHÉ!

b)Ta có:
\(3^{99}>3^{93}=\left(3^3\right)^{21}=27^{21}\)
Vì \(27^{21}>11^{21}\) nên \(3^{99}>27^{21}>11^{21}\) hay \(3^{99}>11^{21}\)
a) Ta có:
19920 < 20020 = 20015.2005
200315 > 200015 = 20015.1015 = 20015.(103)5 = 20015.10005
Vì 19920 < 20015.2005 < 20015.10005 < 200315
=> 19920 < 200315
b) Ta có:
399 = (33)33 = 2733 > 1121
=> 399 > 1121

a, \(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
c, \(x^{10}=1^x\)
\(\Rightarrow x\in\left\{1;0\right\}\)
\(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32=2^4\)
\(\Rightarrow x=4\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
Phần này mk ko bt làm đâu
\(x^{10}=1^x\)
\(\Rightarrow\)\(x^{10}=1\)
\(\Rightarrow x=1\)

Ta có:
\(2^{3^{2^3}}=2^{3^8}=2^{6561}=2^{3.2187}=8^{2187}\)
\(3^{2^{3^2}}=3^{2^9}=3^{512}\)
Ta thấy \(8^{2187}>3^{512}\Rightarrow2^{3^{2^3}}>3^{2^{3^2}}\)
\(2^{3^{2^3}}=2^{3^8}=2^{6561}\)
\(3^{2^{3^2}}=3^{2^9}=3^{512}\)
Tới đây mk chịu để mk suy nghĩ đã!

Ta có:
\(2^{3^{2^3}}=2^{3^8}=2^{6561}=2^{3.2187}=\left(2^3\right)^{2187}=8^{2187}\)
\(3^{2^{3^2}}=3^{2^9}=3^{512}\)
Vì: 8 > 3 và 2187 > 512
\(\Rightarrow8^{2187}>3^{512}\)
\(\Rightarrow2^{3^{2^3}}>3^{2^{3^2}}\)
Vậy: \(2^{3^{2^3}}>3^{2^{3^2}}\)

2^6=64
8^2=64. Vậy 2^6=8^2
5^3=125, 3^5=243. Vì 243>125 nên 5^3<3^5
b: \(3^{30}=\left(3^{10}\right)^3=59049^3\)
\(11^{21}=\left(11^7\right)^3=19487171^3\)
mà 59049<19487171
nên \(3^{30}<11^{21}\)
c: \(72^{45}-72^{44}=72^{44}\cdot\left(72-1\right)=72^{44}\cdot71\)
\(72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)
mà \(72^{44}>72^{43}\)
nên \(72^{45}-72^{44}>72^{44}-72^{43}\)
d: \(2^{500}=\left(2^5\right)^{100}=32^{100}\)
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
mà 32>25
nên \(2^{500}>5^{200}\)
e: \(31^{11}<32^{11}=\left(2^5\right)^{11}=2^{55}\)
\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}>2^{55}\)
Do đó: \(17^{14}>32^{11}\)
f: \(3^{24680}=\left(3^2\right)^{12340}=9^{12340}\)
\(2^{37020}=\left(2^3\right)^{12340}=8^{12340}\)
mà 9>8
nên \(3^{24680}>2^{37020}\)
g: \(2^{1050}=\left(2^7\right)^{150}=128^{150}\)
\(5^{450}=\left(5^3\right)^{150}=125^{150}\)
mà 128>125
nên \(2^{1050}>5^{450}\)
h: \(5^{2n}=\left(5^2\right)^{n}=25^{n}\)
\(2^{5n}=\left(2^5\right)^{n}=32^{n}\)
mà 25<32
nên \(5^{2n}<2^{5n}\)
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