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a/
\(37^{1320}=\left(37^2\right)^{660}=1369^{660}\)
\(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)
\(\Rightarrow1363^{660}>1331^{660}\Rightarrow37^{1320}>11^{1979}\)
b/
\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
\(\Rightarrow27^{11}>81^8\)
d/
\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 9^{21}< 11^{21}\)
e/ \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
g/ \(21^{15}=3^{15}.7^{15}\)
\(27.49^8=3^3.\left(7^2\right)^8=3^3.7^{16}\)
\(\frac{21^{15}}{27.49^8}=\frac{3^{15}.7^{15}}{3^3.7^{16}}=\frac{3^{12}}{7}>1\Rightarrow21^{15}>27.49^8\)
f/ \(199^{20}=\left(199^4\right)^5\)
\(2003^{15}=\left(2003^3\right)^5\)
\(2003^5>1990^5\)
\(\frac{1990^5}{199^4}=\frac{199^5.10^5}{199^4}=199.10^5>1\)
\(\Rightarrow2003^5>1990^5>199^4\Rightarrow2003^{15}>199^{20}\)
a) \(21^{15}=21^{3.5}=\left(21^3\right)^5=9261^5\)
Vì \(9261>27\Rightarrow9261^5>27^5\Rightarrow21^{15}>27^5\)
b) \(15^{12}=\left(3.5\right)^{12}=3^{12}.5^{12}\)
\(81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{4.3}.5^{3.5}=3^{12}.5^{15}\)
Vì \(3^{12}=3^{12}\)mà \(5^{12}< 5^{15}\Rightarrow3^{12}.5^{12}< 3^{12}.5^{15}\Rightarrow15^{12}< 81^3.125^5\)
a) ta có: 3111 > 3015
1714 < 1715 < 3015
=> ...
b) ta có: 291 > 290 = (23)30 = 830 = 530 . 330 =530. 910
535 = 530.55 < 530 . 910
=> ....
c) ta có:2615 < 3015 = 315.1015
1023 = 1015. 1013
mà 1013 > 913 = (33)13 = 329 > 315
=> ....
c) ta có: 2615 < 3015 = 315. 1015
1023 = 1015. 108
mà 108 > 98 = (33)8 = 324 > 315
=> ...
xl bn mk lm sai phần c
a,
\(3^{34}=\left(3^{17}\right)^2=129140163^2\)
\(5^{20}=\left(5^{10}\right)^2=9765625^2\)
Vậy..........
Đùa chút thui
c,\(3^{23}=3^{21}.3^2=\left(3^3\right)^7.9=27^7.9\)
\(5^{15}=\)\(5^{14}.5=\left(5^2\right)^7.5=25^7.5\)
\(27^7>25^7\)và \(9>5\)
nên \(3^{23}>5^{15}\)
b,
5^299 < 5^300 = (5^2)^150 = 25^150
3^501 = (3^3)^167 = 27^167
=> 27^167 > 25^150 => 3^501 > 5^299
a. \(27^{11}=\left(3^3\right)^{11}=3^{3\cdot11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{4\cdot8}=3^{32}\)
Vì 32<33 => 332<333 => 818<2711
b. \(63^{15}=\left(63^5\right)^3=992436543^3\)
\(34^{18}=\left(34^6\right)^3=1544804416^3\)
Vì 992436543<1544804416 nên 9924365433<15448044163 => \(63^{15}< 34^{18}\)