a) Ta có 27¹¹ = (3³)¹¹ = 3³³

81⁸ = (3⁴)⁸ = 3³²

Vì 32 < 33

=> 3³² < 3³³

=> 81⁸ < 27¹¹

b) Ta có 625⁵ = (5⁴)⁵ = 5²⁰

125⁷ = (5³)⁷ = 5²¹

Vì 20 < 21

<=> 5²⁰ < 5²¹

=> 625⁵ < 125⁷

c) Ta có 5³⁶ = (5³)¹² = 125¹²

11²⁴ = (11²)¹² = 121¹²

Vì 125 > 121

<=> 125¹² > 121¹²

<=> 5³⁶ > 11²⁴

a. 27¹¹ và 81⁸

Ta có :

81⁸ = ( 27 ) ^{3 . 8} = 27²⁴

Ta có : 27¹¹ < 27²⁴

=> 27¹¹ < 81⁸

Vậy 27¹¹ < 81⁸

b. 625⁵ và 125⁷

Ta có :

625⁵ = ( 125 )^{5 . 7} = 125³⁵

Ta có : 125³⁵ > 125⁷

=> 625⁵ > 125⁷

Vậy  625⁵ > 125⁷

c. 5³⁶ và 11²⁴

Ta có :

5³⁶ = 5^{3 . 12} = ( 5³ )¹² = 125¹²

11²⁴ = 11 ^{2 . 12} = ( 11² )¹² = 22¹²

Ta có 125¹² < 22¹² 

=> 5²⁶ < 11²⁴

Vậy 5²⁶ < 11²⁴

\(125=5^3\) 

\(25^2=\left(5^2\right)^2=5^{2.2}=5^4\)

\(\left(5^3\right)^2=5^{3.2}=5^6\)

\(125^5=\left(5^3\right)^5=5^{3.5}=5^{15}\)

\(625^4=\left(5^4\right)^4=5^{4.4}=5^{16}\)

\(125.5^2=5^3.5^2=5^{3+2}=5^5\)

125 = 5³

25²  = (5²)² = 5⁴

(5³)² = 5⁶ 

125⁵ = ( 5³)⁵ = 5¹⁵

625⁴ = (5⁴)⁴ = 5¹⁶

125 . 5² = 5³ . 5² = 5⁵

a) 27¹¹ = (3³)¹¹ = 3^3.11 = 3³³

81⁸ = (3⁴)⁸ = 3^4.8 = 3³²

Vì 33³ > 33² => 27¹¹ > 81⁸

b) 625⁵ = (5⁴)⁵ = 5^4.5 = 5²⁰

125⁷ = (5³)⁷ = 5^3.7 = 5²¹

Vì 5²⁰ < 5²¹ => 625⁵ < 125⁷

c) 5³⁶ = 5^3.12 = (5³)¹² = 125¹²

11²⁴ = 11^2.12 = (11²)¹² = 121¹²

Vì 125¹² > 121¹² => 5³⁶ > 11²⁴

5³⁰ x 4¹⁵ = 5³⁰ x 2³⁰ = 10³⁰
mà 10³⁵ >  10³⁰
Vậy 10³⁵ > 5³⁰ x 4¹⁵

5^{2017 >} 119⁷²  k cho mik nha

\(5^{2017}=5^{2016}.5=\left(5^{28}\right)^{72}.5\)

Có \(5^{28}>119\)

\(\Rightarrow\left(5^{28}\right)^{72}>119^{72}\)

\(\Rightarrow5^{2017}>119^{72}\)

a) 5²³ = 5 x 5²²

Do 5 x 5²² < 6 x 5²² 

=> 5²³ < 6 x 5²²

b) 2¹⁶ = 2¹³ x 2³ = 2¹³ x 8

Do 7 x 2¹³ < 2¹³ x 8

=> 7 x 2¹³ < 2¹⁶

c) 21¹⁵ = 3¹⁵ x 7¹⁵

27⁵ x 49⁸ = (3³)⁵ x (7²)⁸ = 3¹⁵ x 7¹⁶

Vì 3¹⁵ x 7¹⁵ < 3¹⁵ x 7¹⁶

=> 21¹⁵ < 27⁵ x 49⁸

Ủng hộ mk nha ^_-

a) 5²³ = 5 x 5²²

Vì 5 < 6 => 5²².5< 6 x 5²²

=>5²³<6x5²²

b) 2¹⁶ = 2¹³ x 2³ = 2¹³ x 8

Vì 7<8 => 7 x 2¹³ < 2¹³ x 8

=> 7 x 2¹³ < 2¹⁶

c) 21¹⁵ = 3¹⁵ x 7¹⁵

27⁵ x 49⁸ = (3³)⁵ x (7²)⁸ = 3¹⁵ x 7¹⁶

Vì 7¹⁵<7¹⁶=>3¹⁵ x 7¹⁵ < 3¹⁵ x 7¹⁶

=> 21¹⁵ < 27⁵ x 49⁸

\(\left(\frac{3}{7}\right)^5.\left(\frac{7}{3}\right)^{-1}.\left(\frac{5}{6}\right)^6.\left(\frac{343}{625}\right)^{-2}\)

\(=\left(\frac{3}{7}\right)^5.\left(\frac{3}{7}\right).\left(\frac{5}{2.3}\right)^6.\left(\frac{5^4}{7^3}\right)^2\)

\(=\frac{3^5}{7^5}.\frac{3}{7}.\frac{5^6}{2^6.3^6}.\frac{5^8}{7^6}=\frac{1}{2^6}.\frac{3^6}{3^6}.\frac{5^{14}}{1}.\frac{1}{7^{12}}=\frac{5^{14}}{2^6.7^{12}}\)

Nó đâu có phải là số nguyên đâu

3/7 x 5/6 +5/6 x3/7

= (3/7x5/6) + (5/6x3/7)

= 15/42 + 15/42  = 30/42 = 5/7

\(2^{5n}=\left(2^5\right)^n=32^n\)

\(5^{2n}=\left(5^2\right)^n=25^n\)

\(32^n>25^n\Rightarrow2^{5n}>5^{2n}\)

Vậy ........

Chúc em học tốt^^

ta c0:

\(2^{5n}=\left(2^5\right)^n=32^n\)

\(5^{2n}=\left(5^2\right)^n=25^n\)

vj 32>25\(\Rightarrow32^n>25^n\)

hay  \(2^{5n}>5^{2n}\)