2^91 và 5^36
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$199^{20}< 200^{20}=100^{20}.2^{20}=100^{20}.(2^5)^4=100^{20}.32^4$
$< 100^{20}.100^4=100^{24}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{151}< 3^{150}\)
Ta có:
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
Mà: \(9>8\)
\(\Rightarrow9^{75}>8^{75}\)
\(\Rightarrow3^{150}>2^{225}\)
\(\Rightarrow3^{151}>2^{225}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(15^{12}=\left(3\cdot5\right)^{12}=3^{12}\cdot5^{12}\)
\(81^3\cdot125^5=\left(3^4\right)^3\cdot\left(5^3\right)^5=3^{12}\cdot5^{15}\)
Mà: \(15>12\)
\(\Rightarrow5^{15}>5^{12}\)
\(\Rightarrow3^{12}\cdot5^{15}>3^{12}\cdot5^{12}\)
\(\Rightarrow81^3\cdot125^5>15^{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^{16}=2^3\cdot2^{13}=8\cdot2^{13}\)
Mà:
\(7< 8\)
\(\Rightarrow7\cdot2^{13}< 8\cdot2^{13}\)
\(\Rightarrow7\cdot2^{13}< 2^{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(5^{23}=5\cdot5^{22}\)
Mà:
\(6>5\)
\(\Rightarrow6\cdot5^{22}>5\cdot5^{22}\)
\(\Rightarrow6\cdot5^{22}>5^{23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(3>2\)
\(\Rightarrow3^{3n}>2^{3n}\) (do n ∈ N)
Vậy: ...
32n và 23n ( n E N*)
Ta có: 32n = (32)n = 9n
23n = (23)n = 8n
Vì 9>8 => 9n > 8n hay 32n > 23n
Vậy 32n > 23n
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$4^{30}=(4^3)^{10}=64^{10}> 48^{10}=(2.24)^{10}=2^{10}.24^{10}> 3.24^{10}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(3^{210}=\left(3^3\right)^{70}=27^{70}\)
\(2^{350}=\left(2^5\right)^{70}=32^{70}\)
Mà: \(32>27\)
\(\Rightarrow32^{70}>27^{70}\)
\(\Rightarrow2^{350}>3^{210}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(32^{60}=\left(2^5\right)^{60}=2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(81^{50}=\left(9^2\right)^{50}=9^{100}\)
Mà: \(8< 9\)
\(\Rightarrow8^{100}< 9^{100}\)
\(\Rightarrow32^{60}< 81^{50}\)
\(2^{91}>2^{90}\)
Ta có:
\(2^{90}=\left(2^5\right)^{18}=32^{18}\)
\(5^{36}=\left(5^2\right)^{18}=25^{18}\)
Mà: \(32>25\)
\(\Rightarrow32^{18}>25^{18}\)
\(\Rightarrow2^{90}>5^{36}\)
\(\Rightarrow2^{91}>5^{36}\)
Ta có:
\(2^{91}>2^{90}=\left(2^5\right)^{18}=32^{18}>25^{18}=\left(5^2\right)^{18}=5^{2.18}=5^{36}\)
Vậy \(2^{91}>5^{36}\).