a)Ta có:\(17^{50}>17^{49}=\left(17^7\right)^7>30^7\)

b)Ta có:\(81^{75}=\left(81^3\right)^{25}=\left(3^{12}\right)^{25}=\left(3^3\right)^{100}=27^{100}< 30^{100}\)

Ta có : 

31/30 > 30/30 

=> 31/30 > 1  ( 1 ) 

22/27 < 27/27

=> 22/27 < 1  ( 2 ) 

Từ ( 1 ) và ( 2 ) 

=>  22/27 < 1 < 31/30

=>  22/27 < 31/30

TK mk nha !!! 

\(a,81^3=\left(9^2\right)^3=9^6\)

Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)

\(b,5^{14}=\left(5^2\right)^7=25^7\)

Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)

\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)

A) 1 - 1974/1975 = 1/1975

1 - 1998/1999 = 1/ 1999

Mặt khác 1/1975 > 1/1999

=> 1974/1975 < 1998/1999

B. 1 - 1999/2005 = 6/2005 

1 - 1993/1999 = 6/1999 

Mặt khác 6/2005 < 6/1999

=> 1999/2005 > 19993/1999

ta có: \(\frac{73}{75}>\frac{73}{79}>\frac{77}{79}\Rightarrow\frac{73}{75}>\frac{77}{79}\)

ta có: \(\frac{53}{100}< \frac{47}{100}\)

ta có: \(\frac{48}{47}>1;\frac{84}{85}< 1\Rightarrow\frac{48}{47}>\frac{84}{85}\)

a, 24^10 < 3^30 + 4^30 + 5^30

b, 2^100 < 5^50 < 3^75.

1,

Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)

\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)

2,

Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)

3,

Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)

\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)

4,

\(\dfrac{48}{47}>1>\dfrac{84}{85}\)

Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)

giúp mình với ạ

cảm ơn ạ

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  > 2017^2337.3 >>> 8000²³³⁷

  7012²⁰¹⁷ < 8000²³³⁷

suy ra:  2017⁷⁰¹² >>> 7012²⁰¹⁷

\(A=1+2+2^2+2^3+......+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+......+2^{101}\)

\(\Rightarrow2A-A=A=2^{101}-1\)

Vì \(2^{101}-1< 2^{101}\)\(\Rightarrow A< B\)

CẢM ƠN NOBITA