\(\left(\frac{3}{8}\right)^5=\frac{243}{8^5}>\frac{125}{8^5}>\frac{125}{9^5}=\frac{125}{\left(3^2\right)^5}=\frac{125}{3^{10}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\left(\frac{5}{243}\right)^3\)

=>\(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)

co cach khac ko

#tranminhkhoi

K bn ak

(3/8)⁵= 3⁵/(2³)⁵=243/2¹⁵>243/3¹⁵>125/3¹⁵=5³/(3⁵)³=(5/3⁵)³=(5/243)³

suy ra (3/8)⁵>(5/243)³

\(243^5=\left(3^5\right)^5=3^{25}\)

\(3.27^8=3.\left(3^3\right)^8=3.3^{24}=3^{25}\)

Vậy\(243^5=3.27^8\)

\(243^5=\left(3^5\right)^5=3^{25}\)

\(3.27^8=3.\left(3^3\right)^8=3.3^{24}=3^{25}\)

Vậy suy ra : \(243^5=3.27^8\)

Như đặc cầu

Ta có: 243⁵=(3⁵)⁵=3^5.5=3²⁵

3.27⁸=3.(3³)⁸=3.3^3.8=3²⁵

=>3²⁵=3²⁵

Vậy 243⁵=3.27⁸

243⁵ = (3⁵)⁵ = 3²⁵

3 . 27⁸ = 3 . (3³)⁸ = 3²⁵

=> 243⁵ = 3 . 27⁸

ta có 243^5=(3^5)^5=3^25  (1)

        3.27^8=3.(3^3)^8=3.3^24=3^25  (2)

        Từ(1)và(2)=>243^5=3.27^8

a/

+ \(\frac{1}{243^6}=\frac{1}{3^6.81^6}=\frac{1}{3^2.3^4.81^6}=\frac{1}{9.81^7}\) (1)

+ \(80< 81\Rightarrow80^7< 81^7\Rightarrow\frac{1}{80^7}>\frac{1}{81^7}\) (2)

+ \(81^7< 9.81^7\Rightarrow\frac{1}{81^7}>\frac{1}{9.81^7}\) (3)

Từ (1) (2) (3) \(\Rightarrow\frac{1}{80^7}>\frac{1}{243^6}\)

b/ Xem lại đề bài

So sánh :

a) 243⁵ và 3 . 27⁸

Ta có : 

243⁵ = ( 3⁵ )⁵ = 3²⁵

3 . 27⁸ = 3 . ( 3³ )⁸ = 3²⁵

Vì 3²⁵ = 3²⁵ nên 243⁵ = 3 . 27⁸

a) 243⁵ = (3⁵)⁵ = 3²⁵

3.27⁸ = 3.(3³)⁸ = 3.3²⁴ = 3²⁵

Vậy 3²⁵ = 3²⁵ . Nên  243⁵ = 3.27⁸ 

b)81³ = (3³)³ = 3⁹

125³ = (5³)³ = 5⁹

81³.125³ = 3⁹.5⁹ = (3.5)⁹ = 15⁹

=> 5¹² < 15⁹ ( 5¹² < 81³.125³ )