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![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$1990^{10}+1990^9=1990^9(1990+1)=1991.1990^9< 1991.1991^9=1991^{10}$
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$10^{10}=(10^2)^5=100^5=(2.50)^5=2^5.50^5=32.50^5< 48.50^5$
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$11^{1979}< 11^{1980}=(11^3)^{660}=1331^{660}$
$37^{1320}=(37^2)^{660}=1369^{660}> 1331^{660}$
$\Rightarrow 11^{1979}< 37^{1320}$
![](https://rs.olm.vn/images/avt/0.png?1311)
b: 99^20=(99^2)^10=9801^10
=>99^20<9999^10
d: 10^10=100^5=4*50^5<48*50^5
e: 1990^10+1990^9
=1990^9(1990+1)
=1990^9*1991
1991^10=1991^9*1991
=>1991^10>1990^9*1991
=>1991^10>1990^10+1990^9
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1.9920và 999910
=(992)10=980110
Vậy 980110<999910 suy ra 9920<999910
Câu 2. 3500và 7300
3500=(35)100=243100
7300=(73)100=343100
Vậy 243100<343100 => 3500<7300
![](https://rs.olm.vn/images/avt/0.png?1311)
f: 11^1979<11^1980=1331^660
37^1320=(37^2)^660=1369^660
1331<1369
=>1331^660<1369^660
=>11^1980<37^1320
=>11^1979<37^1320
g: 10^10=2^10*5^10
48*50^5=2^4*3*2^5*5^10=2^9*3*5^10
2^10<2^9*3
=>2^10*5^10<2^9*3*5^10
=>10^10<48*50^5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(11^{1979}<11^{1980}=\left(11^3\right)^{660}=1331^{660}\)
\(37^{1320}=\left(37^2\right)^{600}=1369^{600}\)
\(1369^{660}>1331^{660}\Rightarrow11^{1979}<37^{1320}\)
b) \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9.1991<1991^9.1991\)
\(\Rightarrow1990^{10}+1990^9<1991^{10}\)
Vậy \(1990^{10}+1990^9<1991^{10}\)
Ta có: \(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)
\(37^{1321}>37^{1320}=\left(37^2\right)^{660}=1369^{660}\)
Vì \(1369^{660}>1331^{660}\)Nên \(11^{1979}< 37^{1321}\)
ta có 11^1979<11^1980=(11^3)^660=1331^660
mà 37^1320=(37^2)^660=1369^660
mà 1331^660>1369^660 vậy 11^1979<37^1320
P/s: ^ là mũ nhé