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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
2^101 > 2^100 = (2^5)^20 = 32^20
5^39 < 5^40 = (5^2)^20 = 25^20
Do 32^20 > 25^20
=> 2^101 > 5^39
\(2^{101}< 2^{100}\Leftrightarrow\left(2^5\right)^{20}=32^{20}\)
\(5^{39}< 5^{40}\Leftrightarrow\left(5^2\right)^{20}=25^{20}\)
Do \(32^{20}>25^{20}\)
nên \(2^{101}>5^{39}\)
Vậy \(2^{101}>5^{39}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 2101 > 2100 = ( 210)10 = 102410 > 62510 = ( 54)10 = 540 > 539
Vậy 2101 > 539
( Bài này sử dụng lũy thừa trung gian bạn nhé )
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
\(...\)
\(\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow M< N\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: A=1+2+2^2+2^3+2^4
=>2A=2+2^2+2^3+2^4+2^5
=>A=2^5-1
=>A=B
b: C=3+3^2+...+3^100
=>3C=3^2+3^3+...+3^101
=>2C=3^101-3
=>\(C=\dfrac{3^{101}-3}{2}\)
=>C=D
Ta có:
\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)
\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 100+101/101+102
= 100/101+102 + 101/101+102
Vì 100/101>100/101+102
101/102 > 101/101+102
=>100/101+101/102 > 100+101/101+102
![](https://rs.olm.vn/images/avt/0.png?1311)
101/99=1-2/99
81/79=1-2/79
mà -2/99>-2/79
nên 101/99>81/79
\(2^{101}< 2^{100}\Leftrightarrow\left(2^5\right)^{20}=32^{20}\)
\(5^{39}< 5^{40}\Leftrightarrow\left(5^2\right)^{20}=25^{20}\)
Ta thấy: \(32^{20}>25^{20}\)
\(\Rightarrow2^{101}>5^{39}\)