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\(2017^{7012}>2017^{6051}=\left(2017^3\right)^{2017}\)
Mà \(2017^3>2017\)
\(\Rightarrow\)\(2017^{2012}>7012^{2017}\)
\(A=\frac{2017^{99}}{2017^{100}-2}\)
=> \(2017A=\frac{2017^{100}}{2017^{100}-2}=\frac{2017^{100}-2+2}{2017^{100}-2}=1+\frac{2}{2017^{100}-2}\)
\(B=\frac{2017^{100}}{2017^{101}-2}\)
=>\(2017B=\frac{2017^{101}}{2017^{101}-2}=\frac{2017^{101}-2+2}{2017^{101}-2}=1+\frac{2}{2017^{101}-2}\)
Do \(\frac{2}{2017^{100}-2}>\frac{2}{2017^{101}-2}\)
Nên 2017A > 2017B
Vậy A > B
Ta có :
20177012 > 20176051 = (20173)2017
Mà 20173 > 7012
=> 20172012 > 70122017
\(2013^{3012}\)và \(3012^{2013}\)
\(2013^{3012}=\left(3.671\right)^{3012}\)
\(3012^{2013}=\left(3.1004\right)^{2013}\)
Ta thấy : \(\left(3.671\right)^{3012}>\left(3.1004\right)^{2013}\)
\(\Rightarrow2013^{3012}>3012^{2013}\)
A=(1.1-2.2)+(3.3-4.4)+...+(99.99-100.100)+101.101
A= (-3)+(-7)+...+(-199)+101.101
A=-[(199+3).50:2]+101.101
A= -5050+101.101
A=101.(-50)+101.101=(-50.101).101=510050
a)A=3^0+3^1+3^2+3^3+...+3^2012
A=1+3+3^2+3^3+..+3^2012
3A=3+3^2+3^3+3^4+..+3^2013
3A-A=3+3^2+3^3+3^4+..+3^2013-1-3-3^2-3^3-...-3^2012
2A=3^2013-1
A=\(\frac{3^{2013}-1}{2}\)
B=3^2013
=> A>B
b) A=1+5+5^2+5^3+..+5^99+5^100
5A=5+5^2+5^3+5^4+...+5^100+5^101
5A-A=5+5^2+5^3+5^4+..+5^100+5^101-1-5-5^2-5^3-..-5^99-5^100
4A=5^101-1
A=\(\frac{5^{101}-1}{4}\)
B=5^101/4
=> A<B
a )
2100+2100= 2100(1+1) =2100.2 = 2100+1= 2101
b)
3100+3100 = 3100(1+1) = 2.3100
3101= 3100.3
ta thấy 3. 3100 > 2.3100 Vậy 3101 > 3100+3100
c) 20177012 > 20172337.3 >>> 80002337
70122017 < 80002337
suy ra: 20177012 >>> 70122017