Ta có: \(3^{2010}=3^{10}\cdot3^{2000}=3^{10}\cdot9^{1000}\)

\(2^{3010}=2^{10}\cdot2^{3000}=2^{10}\cdot8^{1000}\)

Xét thấy \(3^{10}>2^{10};9^{1000}>8^{1000}\)

\(\Rightarrow3^{10}\cdot9^{1000}>2^{10}\cdot8^{1000}\)

Vậy \(3^{2010}>2^{3010}\)

a)  1024 9 = ( 2 10 ) 9 = 2 90 < 2 100

b)  6 . 5 29 > 5 . 5 29 = 5 30

c) 10 30 = ( 10 3 ) 10 = 1000 10 ; 2 100 = ( 2 10 ) 10 = 1024 10   n ê n   10 30 < 2 100 .

a) >

b) <

c) <

d) <

a)>

b)<

c)<

d)<

a) Cách 1:  2 100 = 2 10 10 = 1024 10 > 1024 9

Cách 2:  1024 9 = 2 10 9 =  2 90  <  2 100

b)  6 . 5 29  >  5 . 5 29  =  5 30

c)  2 98 =  2 2 49 =  4 49  <  9 49

d)  10 30 =  10 3 10 = 1000 10 ;  2 100 =  2 10 10  = 1024 10 nên  10 30  <  2 100

\(\frac{339}{322}\)và \(\frac{338}{321}\)

Ta có : \(\frac{339}{322}-1=\frac{17}{322};\frac{338}{321}-1=\frac{17}{321}\).

Vì \(\frac{17}{322}< \frac{17}{321}\)nên \(\frac{339}{322}< \frac{338}{321}\).

\(\frac{2017}{2014}\)và \(\frac{2018}{2015}\)

Ta có : \(\frac{2017}{2014}-1=\frac{3}{2014};\frac{2018}{2015}-1=\frac{3}{2015}\)

Vì \(\frac{3}{2014}>\frac{3}{2015}\)nên \(\frac{2017}{2014}>\frac{2018}{2015}\).

\(\frac{511}{514}\)và \(\frac{513}{516}\)

Ta có : \(1-\frac{511}{514}=\frac{3}{514};1-\frac{513}{516}=\frac{3}{516}\)

Vì \(\frac{3}{514}>\frac{3}{516}\)nên \(\frac{511}{514}< \frac{513}{516}\).

\(\frac{3005}{3000}\)và \(\frac{3010}{3005}\)

Ta có : \(\frac{3005}{3000}-1=\frac{5}{3000};\frac{3010}{3005}-1=\frac{5}{3005}\)

Vì \(\frac{5}{3000}>\frac{3}{3005}\)nên \(\frac{3005}{3000}>\frac{3010}{3005}\).

~ Chúc bn hok tốt ~

`@` `\text {Ans}`

`\downarrow`

`2^100` và `3^50`

Ta có:

\(2^{100}=\left(2^4\right)^{25}=16^{25}\)

\(3^{50}=\left(3^2\right)^{25}=9^{25}\)

Vì `16 > 9 =>`\(16^{25}>9^{25}\Rightarrow2^{100}>3^{50}\)

Vậy, `2^100 > 3^50` `.`

Ta có: 2¹⁰⁰=2³¹.2⁶⁹

  = 231 . 263 . 26

                 = 231 . ( 29 )7 . ( 22)3

                  = 231 . 5127 . 43 

Lại có : 1031 = 231 . 531

                          = 231 . 528 . 53

                                 = 231 . ( 54) 7 . 53

                        = 231 . 6257 . 53 

=>231 . 6257 . 53 > 231 . 3127 . 53 > 231 . 3127 . 43

<=> 2¹⁰⁰<10³¹

2¹⁰⁰ < 10³¹

Ta có:

\(2^{200}.2^{100}=\left(2^2\right)^{100}.2^{100}=4^{100}.2^{100}=\left(4.2\right)^{100}=8^{100}\)

\(3^{100}.3^{100}=\left(3.3\right)^{100}=9^{100}\)

Vì \(8< 9\) nên \(8^{100}< 9^{100}\)

Vậy \(2^{200}.2^{100}< 3^{100}.3^{100}\)

\(#WendyDang\)

\(2^{200}\cdot2^{100}=2^{300}=(2^3)^{100}=8^{100}\\3^{100}\cdot3^{100}=(3\cdot3)^{100}=9^{100}\)

Vì \(8< 9\) nên \(8^{100}< 9^{100}\)

hay \(2^{200}\cdot2^{100}< 3^{100}\cdot3^{100}\)

\(A=8^{200}=\left(2^3\right)^{200}=2^{600}=2^{100}\cdot2^{500}\\ B=2^{100}\cdot9^{150}=2^{100}\cdot\left(3^2\right)^{150}=2^{100}\cdot3^{300}\\ 2^{500}=32^{100};3^{300}=27^{100}\\ 32^{100}>27^{100}\Rightarrow2^{500}>3^{300}\\ \Rightarrow A>B\)