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\(a,3^{39}=\left(3^3\right)^{13}=9^{13}< 11^{13}< 11^{21}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7< 3125^7=\left(5^5\right)^7=5^{35}\)
(x-1)2020=(x-1)2022
=>(x-1)2020-(x-1)2022=0
=>(x-1)2020-(x-1)2020.(x-1)2=0
=>(x-1)2020(1-(x-1)2=0
=>(x-1)2020=0 hoặc 1-(x-1)2=0
=>x=1 hoặc x=2.
Bài 2
a,2105 và 545
2105=(27)15=12815
545=(53)15=12515
Vì 12815>12515 nên 2105>545.
b,
554 và 381
554=(56)9=156259
381=(39)9=196839
Vì 156259<196839 nên 554<381
Bài 1 :
\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)
\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)
\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
291 và 535
291 = (213)7 = 81927
535 = (55)7 = 31257
Vì 81927 > 31257 => 291 > 535
Vậy 291 > 535
a. \(5^{127}=5.5^{126}=5.125^{72}>119^{72}\)
\(\Rightarrow5^{217}>119^{72}\)
b. \(2^{1000}=\left(2^5\right)^{200}=32^{200}\)
\(5^{400}=\left(5^2\right)^{200}=25^{200}\)
\(\Rightarrow2^{1000}>5^{400}\)
c. \(9^{12}=\left(3^2\right)^{12}=3^{24}\)
\(27^7=\left(3^3\right)^7=3^{21}\)
\(\Rightarrow9^{12}>27^7\)
d. \(125^{80}=\left(5^3\right)^{80}=5^{240}\)
\(25^{118}=\left(5^2\right)^{118}=5^{236}\)
\(\Rightarrow125^{80}>25^{118}\)
e. \(5^{40}=\left(5^4\right)^{10}=625^{10}\)
\(\Rightarrow5^{40}>620^{10}\)
f. \(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
\(\Rightarrow27^{11}>81^8\)
a) \(5^{48}=\left(5^4\right)^{12}=625^{12}\)
\(2^{108}=\left(2^9\right)^{12}=512^{12}\)
Do \(625>512\Rightarrow625^{12}>512^{12}\) \(\Rightarrow5^{48}>2^{108}\) (1)
Lại có: \(108>105\Rightarrow2^{108}>2^{105}\) (2)
Từ (1) và (2) \(\Rightarrow5^{48}>2^{105}\)
b) \(2^{50}=\left(2^5\right)^{10}=32^{10}\)
Do \(33>32\Rightarrow33^{10}>32^{10}\)
Vậy \(33^{10}>2^{50}\)
c) Do \(513>512\Rightarrow513^{100}>512^{100}\) (1)
\(512^{100}=\left(2^9\right)^{100}=2^{900}\) \(=2^{10.90}=\left(2^{10}\right)^{90}=1024^{90}\) (2)
Do \(1024>1023\Rightarrow1024^{90}>1023^{90}\) (3)
Từ (1), (2) và (3) \(\Rightarrow513^{100}>1023^{90}\)
a/
\(27^{81}=\left(3^3\right)^{81}=3^{241}\)
\(81^{27}=\left(3^4\right)^{27}=3^{108}\)
\(\Rightarrow27^{81}=3^{241}>3^{108}=81^{27}\)
b/
\(5^{60}=\left(5^3\right)^{20}=125^{20}\)
\(7^{40}=\left(7^2\right)^{20}=49^{20}\)
\(\Rightarrow5^{60}=125^{20}>49^{20}=7^{40}\)
c/
\(11^{102}=\left(11^2\right)^{51}=121^{51}>121^{50}>99^{50}\)
d. So sánh a=12^34567 với b=(12^5)^12=12^60 => a>b
so sánh b=(12^5)^12 với c=34567^12 => b>c
Vậy a>c.
\(a,81^3=\left(9^2\right)^3=9^6\)
Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)
\(b,5^{14}=\left(5^2\right)^7=25^7\)
Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)
\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)