a) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

a) Ta có :

2727>2726=(272)13=72913>243132727>2726=(272)13=72913>24313

⇒2727>24313⇒2727>24313

⇒−2727<−24313⇒−2727<−24313

⇒(−27)27<(−243)13⇒(−27)27<(−243)13

b) (18)25>(18)26=(182)13=(164)13>(1128)13(81​)25>(81​)26=(821​)13=(641​)13>(1281​)13

⇒(18)25>(1128)13⇒(81​)25>(1281​)13

⇒(−18)25<(−1128)13⇒(−81​)25<(−1281​)13

c) 450=(45)10=102410450=(45)10=102410

830=(83)10=51210<102410830=(83)10=51210<102410

⇒450>830⇒450>830

d) (19)17<(19)12<(127)12(91​)17<(91​)12<(271​)12

⇒(19)17<(127)12⇒(91​)17<(271​)12

a) 27⁵ và 243³

Ta có :

27⁵ = ( 3³ )⁵ = 3¹⁵

243³ = ( 3⁵ )³ = 3¹⁵

Vì 3¹⁵ = 3¹⁵ Nên 27⁵ = 243³

b) 2³⁰⁰ và 3²⁰⁰

Ta có :

2³⁰⁰ = ( 2³ )¹⁰⁰ = 8¹⁰⁰

3²⁰⁰ = ( 3² )¹⁰⁰ = 9¹⁰⁰

Vì 8¹⁰⁰ < 9¹⁰⁰ Nên 2³⁰⁰ < 3²⁰⁰

c) 125⁵ và 25⁷

Ta có : 

125⁵ = ( 5³ )⁵ = 5¹⁵

25⁷ = ( 5² )⁷ = 5¹⁴

Vì 5¹⁵ > 5¹⁴ Nên 125⁵ > 27⁷

d) 9²⁰ và 27¹³

Ta có : 

9²⁰ = ( 3² )²⁰ = 3⁴⁰

27¹³ = ( 3³ )¹³ = 3³⁹

Vì 3⁴⁰ > 3³⁹ Nên 9²⁰ > 27¹³

e) 3⁵⁴ và 2⁸¹

Ta có : 

3⁵⁴ = ( 3² )²⁷ = 9²⁷

2⁸¹ = ( 2³ )²⁷ = 8²⁷

Vì 9²⁷ > 8²⁷ Nên 3⁵⁴ > 2⁸¹

g) 10³⁰ và 2¹⁰⁰

Ta có :

10³⁰ = ( 10³ )¹⁰ = 1000¹⁰

2¹⁰⁰ = ( 2¹⁰ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ Nên 10³⁰ < 2¹⁰⁰

A/ 27^5 =243^3

B/2^300<3^200

C/125^5>25^7

D/9^20>27^13

E/3^54>2^81

G/10^30<2^100

Chắc chắn 100% 

\(\left(\frac{1}{27}\right)^{10}\&\left(\frac{1}{243}\right)^7\)

\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)

\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)

Vậy \(\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)

Ta có : 

\(\left(\frac{1}{27}\right)^{10}=\left(\frac{1}{3^3}\right)^{10}=\frac{1}{3^{30}}\)

\(\left(\frac{1}{243}\right)^7=\left(\frac{1}{3^5}\right)^7=\frac{1}{3^{35}}\)

Do :     \(\frac{1}{3^{30}}>\frac{1}{3^{35}}\left(3^{30}< 3^{35}\right)\)

\(\Rightarrow\left(\frac{1}{27}\right)^{10}>\left(\frac{1}{243}\right)^7\)

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

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

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

\(27^{45}=\left(3^3\right)^{45}=3^{135}\\ 243^{27}=\left(3^5\right)^{27}=3^{135}\)

Vậy \(27^{45}=243^{27}\)

27^5 = (3^3)^5 = 3^3.5 = 3^15

243^3 = (3^5)^3 = 3^5.3 = 3^15 

Vậy 27^5 = 243^3

Giải

\(\left(-27\right)^{27}=\left(-3^3\right)^{27}=\left(-3\right)^{81}\)

\(\left(-243\right)^{13}=\left(-3^5\right)^{13}=\left(-3\right)^{63}\)

=>\(\left(-27\right)^{27}>\left(-243\right)^{13}\)

Bạn học tốt!^^

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

27⁵ và 243³

27⁵=(3³)⁵=3¹⁵

243³=(3⁵)³=3¹⁵

=>27⁵=243³

27⁵=(3³)⁵=3¹⁵

243³=(3⁵)³=3¹⁵

Vậy 27⁵=243³

27^5 = (3^3)^5 = 3^3.5 = 3^15

243^3 = (3^5)^3 = 3^5.3 = 3^15 

Vậy 27^5 = 243^3