a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

a ) 

\(3^{400}=\left(3^4\right)^{100}=81^{100}\)

\(9^{200}=\left(9^2\right)^{100}=81^{100}\)

Ta có : \(81^{100}=81^{100}\)

\(\Rightarrow3^{400}=9^{200}\)

b ) 

\(2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{223}=\left(3^2\right)^{111}=9^{111}\)

Ta cos : \(8^{111}< 9^{111}\)

\(\Rightarrow2^{332}< 3^{223}\).

a) 9²⁰⁰  = 3⁴⁰⁰ => 3⁴⁰⁰ < 9²⁰⁰

b) 2³³² = 4. 8¹¹¹

    3²²³ = 3.9¹¹¹

^=> 2³³² < 3²²³

a) Ta có: 9^200 = (3^2)^200 = 3^400

Vì 3^400 = 3^400

=> 3^400 = 9^200

b) Ta có: 2^332 < 2^333 = (2^3)^111 = 8^111

3^223 > 3^222 = (3^2)^111  = 9^111

Vì 8^111 < 9^111 nên 2^332 < 3^223

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

\(d,10^{20}=\left(10^2\right)^{10}=100^{10}\\ Vì:100^{10}>20^{10}\left(Do:100>20\right)\\ \Rightarrow10^{20}>20^{10}\)

2332< 2333=(23)111=8111

3223>3222=(32)111=9111

mà 8<9

=> 2332<3223

đúng tk cho mik

Ta có: 2³³² = 2³³⁰⁺² = 2³³⁰ .2² = (2³)¹¹⁰.4= 8¹¹⁰ . 4

          3²²³ = 3²²⁰⁺³ = 3²²⁰ . 3³ =(3²)¹¹⁰ . 27=9¹¹⁰ . 27

Vì 8<9 , 4<27  => 8¹¹⁰< 9¹¹⁰

                       => 8¹¹⁰ . 4 < 9¹¹⁰ . 27

                       => 2³³² < 3²²³

Ta có:2³³²<2³³³= (2³)¹¹¹ =8¹¹¹

           3²²³>3²²²= (3²)¹¹¹ =9¹¹¹

Vì 8¹¹¹<9¹¹¹nên

2³³²<8¹¹¹<9¹¹¹<3²²³   => 2³³²< 3²²³

Vậy 2³³²< 3²²³ .

2³³²< 2³³³=(2³)¹¹¹=8¹¹¹

3²²³>3²²²=(3²)¹¹¹=9¹¹¹

mà 8<9

=> 2³³²<3²²³

2^332 < 2^333 =  (2^3)^111 = 8^111 < 9^111 = (3^2)^111 = 3^222 < 3^223