\(A=\dfrac{10^{2017}+1}{10^{2018}+1}\)

=>\(10A=\dfrac{10^{2018}+1+9}{10^{2018}+1}=1+\dfrac{9}{10^{2018}+1}\)

\(B=\dfrac{10^{2018}+1}{10^{2019}+1}\)

=>\(10B=\dfrac{10^{2019}+1+9}{10^{2019}+1}=1+\dfrac{9}{10^{2019}+1}\)

Do đó:\(10B< 10A\)=>\(B< A\)

\(A=\dfrac{10^{2017}+1}{10^{2018}+1}\)

\(10A=\dfrac{10\left(10^{2017}+1\right)}{10^{2018}+1}=\dfrac{10^{2018}+10}{10^{2018}+1}=\dfrac{10^{2018}+1+9}{10^{2018}+1}=\dfrac{10^{2018}+1}{10^{2018}+1}+\dfrac{9}{10^{2018}+1}=1+\dfrac{9}{10^{2018}+1}\)\(B=\dfrac{10^{2018}+1}{10^{2019}+1}\)

\(10B=\dfrac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\dfrac{10^{2019}+10}{10^{2019}+1}=\dfrac{10^{2019}+1+9}{10^{2019}+1}=\dfrac{10^{2019}+1}{10^{2019}+1}+\dfrac{9}{10^{2019}+1}=1+\dfrac{9}{10^{2019}+1}\)Vì \(1+\dfrac{9}{10^{2018}+1}>1+\dfrac{9}{10^{2019}+1}\)

Nên \(10A>10B\)

Nên \(A>B\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

`3^12` và `5^8`

\(3^{12}=\left(3^3\right)^4=9^4\)

\(5^8=\left(5^2\right)^4=25^4\)

Vì `9 < 25` `=> 25^4 > 9^4`

`=> 3^12 > 5^8`

Vậy, `3^12 > 5^8`

`b)`

`(0,6)^9` và `(-0,9)^6`

\(\left(0,6\right)^9=\left(0,6^3\right)^3=\left(0,216\right)^3\)

\(\left(-0,9\right)^6=\left[\left(-0,9\right)^2\right]^3=\left(0,81\right)^3\)

Vì `0,81 > 0,216 => (0,81)^3 > (0,216)^3`

`=> (0,6)^9 < (-0,9)^6`

Vậy, `(0,6)^9<(-0,9)^6`

1.a) Có 3¹² = 3^3.4 = 27⁴ ;

5⁸ = 5^2.4 = 25⁴ 

Dễ thấy 27⁴ > 25⁴ nên 3¹² > 5⁸

b) Có  \(0,6^9=\dfrac{6^9}{10^9}=\dfrac{6^{3.3}}{10^9}=\dfrac{216^3}{10^9}\) 

mà \(\left(-0,9\right)^6=0,9^6=\dfrac{9^6}{10^6}=\dfrac{9^6.10^3}{10^9}=\dfrac{9^{2.3}.10^3}{10^9}=\dfrac{81^3.10^3}{10^9}=\dfrac{810^3}{10^9}\)

Dễ thấy \(\dfrac{216^3}{10^9}< \dfrac{810^3}{10^9}\Rightarrow0,6^9< \left(-0,9\right)^6\)

a: 27^4=(3^3)^4=3^12<3^18

b: 49^4=(7^2)^4=7^8

c: 9^16=(9^2)^8=81^8>27^8

a) \(\dfrac{5}{9}< \dfrac{7}{9}\)

b) \(\dfrac{7}{6}>\dfrac{6}{6}\)

c) \(\dfrac{3}{14}< \dfrac{5}{14}\)

d) \(\dfrac{5}{8}< \dfrac{9}{8}\)

a, >

b, <

c, >

A 3/7 > 2/5 

B 5/9 < 5/8 

C 8/7 > 7/8

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