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\(a,81^3=\left(9^2\right)^3=9^6\)
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 \(10^{30}< 2^{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 9\(^{10}\)=\(\left(3^2\right)^{10}\)=\(3^{20}\)(1)
\(27^{13}=\left(3^3\right)^{13}\)=\(3^{39}\)(2)
Từ (1) và (2) => \(27^{13}>9^{10}\)
910 và 2713
910 = ( 32 )10 = 320
2713 = ( 33 )13 = 339
Vì 20 < 39 nên 910 < 2713
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{27}\right)^{10}\&\left(\frac{1}{243}\right)^7\)
\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)
\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)
Vậy \(\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)
Ta có :
\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)
\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)
Do : \(\frac{1}{3^{30}}>\frac{1}{3^{35}}\left(3^{30}< 3^{35}\right)\)
\(\Rightarrow\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`M=(10^25+1)/(10^26+1)`
`=>10M=(10^26+10)/(10^26+1)=1+9/(10^26+1)``
`CMTT:10N=1+9/(10^27+1)`
Vì `1/(10^26+1)>1/(10^27+1)`
`=>9/(10^26+1)>9/(10^27+1)`
`=>1+9/(10^26+1)>1+9/(10^27+1)`
`=>10M>10N=>M>N`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=10^{25}+\frac{1}{10^{26}}+1=1\cdot10^{25}\)
\(B=10^{26}+\frac{1}{10^{27}}+1=1\cdot10^{26}\)
\(1\cdot10^{25}< 1\cdot10^{26}\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(21^{30}=3^{30}.7^{30}\)
\(27^{10}.49^{16}=\left(3^3\right)^{10}.\left(7^2\right)^{16}=3^{30}.7^{32}\)
Vì 7^32>7^30=> 21^30<27^10.49^16