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a, 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975 nên 2225 < 3150
b, 334 > 330 = (33)10 = 2710
521 > 520 = (52)10 = 2510
Vì 2710 > 2510 => 330 > 520 => 334 > 521
c, 321 > 320 = (32)10 = 910
231 > 230 = (23)10 = 810
Vì 910 > 810 => 321 > 231
d, 291 > 290 = (25)18 = 3218
535 < 536 = (52)18 = 2518
Vì 3218 > 2518 => 291 > 535
e, 9920 = (992)10 = 980110 < 999910
f, 128.912 = 38.48.324 = 332.212
1816 = 216.916 = 216.332
Vì 332 . 212 < 216.332 => 128.912 < 1816
g, 7520 = 2520.320 = 540.320
4510.530 = 510.910.530 = 540.320
Vậy 7520 = 4510.530
\(99^{20}< 9999^{10}\)
\(54^4< 21^{12}\)
\(71^5< 17^{20}\)
\(2^{30}+3^{20}+4^{30}>3\times24^{10}\)
a. \(107^{50}< 73^{75}\)
b. \(54^4< 21^{12}\)
c. \(2^{91}>5^{35}\)
d. \(3^{500}>7^{300}\)
e. \(122^{26}< 10^{51}\)
g. \(124^{30}>625^{23}\)
2^24 = (2^3)^8 = 8^8
3^16 = (3^2)^8 = 9^8
Vì 8^8 < 9^8 => 2^24 < 3^16
99^20 = 99^10 . 99^10 < 99^10 . 101^110 = (99.101)^10 = 9999^10
=> 99^20 < 9999^10
2^91 = (2^13)^7 = 8192^7
5^35 = (5^5)^7 = 3125^7
Vì 8192^7 > 3125^7 => 2^91 > 5^35
k mk nha
Ta có:\(2^{36}\)và \(3^{27}\)
\(2^{36}=\left(2^4\right)^9=16^9\)
\(3^{27}=\left(3^3\right)^9=27^9\)
Vì \(16< 27\Rightarrow16^9< 27^9\)
Vậy....
b,\(9^{20}\)và \(9999^{10}\)
\(9^{20}=\left(9^2\right)^{10}=81^{10}\)
\(9999^{10}\)
Vì \(81< 9999\Rightarrow81^{10}< 9999^{10}\)
Vậy ...
c,\(54^4\)
\(21^{12}=\left(21^3\right)^4=9261^4\)
Vì \(54< 9261\Rightarrow54^4< 9261^4\)
Vậy...
\(26^{14}>25^{14}=\left(5^2\right)^{14}=5^{28}\)
\(5^{30}=\left(5^3\right)^{10}=125^{10}>124^{10}\)
\(4^{21}=\left(4^3\right)^7=64^7>64^2\)
\(27^{16}.16^9=\left(3^3\right)^{16}.\left(4^2\right)^9=3^{48}.4^{18}>12^{18}=3^{18}.4^{18}\)
\(31^{11}16^{14}=\left(2^4\right)^{14}=2^{56}\)
\(2^{56}>2^{55}\) => \(17^{14}>31^{11}\)
Các bài khác làm tương tự
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, 2225 = 215.15= ( 215)15 = 3276815
3150 = 310.15 = ( 310)15 = 5904915
Dễ thấy 32768 < 59049 nên 2225 < 3150