a 444³³³³  = 333⁴⁴⁴⁴

b 348⁴  < 4³⁶³

c 1990¹⁰  +1990⁹  > 1991¹⁰

d 2²⁰⁰⁴  > 5⁸⁹¹

^e 10³¹ > 2¹⁰⁰

^{k mik nha bn !!! mình làm nanh nhất !}

Bạn Trịnh Quang phần a bn sai rồi. 444³³³³< 333⁴⁴⁴⁴ nha bạn

a, Ta có :

\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

bạn so sánh nha :)

b,

T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)

tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk

b)2^300=(2^3)^100=8^100

  3^200=(3^2)^100=9^100

vi 8<9nen 2^300<3^200

Ta có \(3^{21}=\left(3^3\right)^7=27^7\)

\(2^{31}=2147483648\)

Mà \(27>2_{ }\)\(\Rightarrow3^{21}>2^{31}\)

c)

\(32^9>18^{13}\)(chứng minh tương tự) 

b) (2^3)^8=8^8

(3^2)^8=9^8

=>2^24<3^16(8^8<9^8)

Theo mình :
2³³⁵ và 3²²⁵
2³³³<2³³⁵ ; 3²²²<3²²⁵
2³³³=(2³)¹¹¹= 8¹¹¹
3²²²= (3²)¹¹¹= 9¹¹¹
8¹¹¹<9¹¹¹
=> 8¹¹¹<2³³⁵<9¹¹¹<3²²⁵
Vậy : 2³³⁵ <3²²⁵ 

a) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

b) Ta có: \(2^{31}=\left(2\frac{31}{21}\right)^{21}=2,7822^{21}< 3^{21}\Rightarrow2^{31}< 3^{21}\)

c) Ta có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\)

\(2^{30}=\left(2^3\right)^{10}=8^{10}\)

\(4^{30}=\left(4^3\right)^{10}=64^{10}\)

Lại có: \(3.24^{10}=2.24^{10}+24^{10}\Rightarrow24^{10}< 27^{10}\left(1\right)\)

\(2.24^{10}< 48^{10}< 64^{10}\left(2\right)\)

Từ 1,2 => \(24^{10}+2.24^{10}< 27^{10}+64^{10}\Rightarrow3.24^{10}< 8^{10}+27^{10}+64^{10}\)

\(\Rightarrow3.24^{10}< 3^{30}+2^{30}+4^{30}\)

2¹⁰⁰=(2¹⁰)¹⁰=1024¹⁰

10³⁰=(10³)¹⁰=1000¹⁰

vì 1024>1000 nên 1024¹⁰>1000¹⁰

hay 2¹⁰⁰>10³⁰

1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)

2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)

c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)

d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)

\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)

4.

a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)

Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)

\(\Rightarrow3^{4000}=9^{2000}\)

c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)

\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)

Ta thấy \(4.8^{110}< 9.9^{110}\)

Vậy \(2^{332}< 3^{223}\)