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a) Ta có: 2500 = (25)100 = 32100
5200= (52)100= 25100
Vì 32100>25100 => 2500 > 5200
![](https://rs.olm.vn/images/avt/0.png?1311)
2500=(25)100=32100
5200=(52)100=25100
vì 32100 > 25100 \(\Rightarrow\)2500 > 5200
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 2500 = 25.100 = (25)100 = 32100
5200 = 52.100 = (52)100 = 25100
Vì 32 > 25 nên 32100 > 25100 nên 2500 > 5200
Vậy 2500 > 5200
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{200}-1\right)\)
\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{200}\right)\)
\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{199}{200}\)
\(-A=\frac{1}{200}\)
\(A=\frac{-1}{200}>\frac{-1}{199}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
Đặt a - b = x ; b - c = y ; c - a = z
=> x + y + z = 0
Ta có :
\(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2.\left(\frac{x+y+z}{xyz}\right)\)
=> \(N=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)( Vì x + y + z = 0 )
Vậy ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(2^{500}=\left(2^5\right)^{100}=32^{100}\)
Vậy \(5^{200}< 2^{500}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2^{500}=2^{5.100}=32^{100}\)
\(5^{200}=5^{2.100}=25^{100}\)
Vì \(32^{100}>25^{100}\) nên \(2^{500}>5^{200}\)
Ta có ; 2500 = (25)100 = 32100
5200 = (52)100 = 25100
Vậy 2500 > 5200
Ta có:
\(2^{500}=\left(2^5\right)^{100}=32^{100}\)
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(\Rightarrow32^{100}>25^{100}\Rightarrow2^{500}>5^{200}\)
Ai k mik mik k lại nha