a) ta có: 3¹⁰⁰ = (3²)⁵⁰ = 9⁵⁰

b) ta có: 3³⁰ = (3³)¹⁰ = 27¹⁰ > 8¹⁰

c) ta có: 36.6⁷ = 6².6⁷ = 6⁹ 

Lại có: 4³³ > 4²⁷ = (4³)⁹ = 64⁹ > 6⁹

=> 4³³>36.6⁷

\(a,\)\(3^{100}\)\(=3^{2.50}\)=\(\left(3^2\right)\)\(^{50}\)\(=9^{50}\)

\(\Rightarrow\)\(3^{100}\)= \(9^{50}\)

b,\(3^{30}\)\(=3^{3.10}\)\(=\left(3^3\right)\)\(^{10}\)\(=27^{10}\)

\(\Rightarrow\)\(3^{30}\)\(>8^{10}\)

\(9^{100}và3^{200}=3^{200}và3^{200}\\ \Rightarrow3^{200}=3^{200}\\ \Rightarrow9^{100}=3^{200}.\\ 5^{23}và125^3=5^{23}và5^9\\ \Rightarrow5^{23}>5^9\\ \Rightarrow5^{23}>5^3.\)

9¹⁰⁰ = (3²)¹⁰⁰ = 3²⁰⁰

Vậy 9¹⁰⁰ = 3²⁰⁰

------------

125³ = (5³)³ = 5⁹

Do 23 > 9 nên 5²³ > 5⁹

Vậy 5²³ > 125³

ss 9 mũ 100 và 3 mũ 200 thì đổi 9 mũ 100 ra thành 3 mũ 2 mũ 100 đi

Ko ghi đề

\(2A=2+2^2+...+2^{101}\\ 2A-A=2^{101}-1\\ =>A=2^{101}-1\)

Mấy cái khác cg lm như v (b thì 3b)

Nhớ đúng mk nhá

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

\(5^{20}=\left(5^2\right)^{10}=25^{10}\)

Suy ra: 2⁵⁰ > 5²⁰

b)

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

Suy ra: 99¹⁰⁰ > 81¹⁰⁰

\(5^{202}=\left(5^2\right)^{101}=25^{101}\)

\(2^{505}=\left(2^5\right)^{101}=32^{101}\)

Suy ra: 5²⁰² < 2⁵⁰⁵

nhanh lên các bn m còn 30 p nữa sắp phải nộp rùi

ai trả lời nhanh nhất m k cho

Bài 1:

a: \(10^{10}=\left(2\cdot5\right)^{10}=2^{10}\cdot5^{10}=2^9\cdot5^{10}\cdot2\)

\(48\cdot50^5=2^4\cdot3\cdot\left(2\cdot5^2\right)^5=2^4\cdot3\cdot2^5\cdot5^{10}=2^9\cdot5^{10}\cdot3\)

mà 2<3

nên \(10^{10}<48\cdot50^5\)

b: \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9\cdot1991\)

\(1991^{10}=1991^9\cdot1991\)

mà 1990<1991

nên \(1990^{10}+1990^9<1991^{10}\)

c: \(107^{50}<108^{50}=\left(2^2\cdot3^3\right)^{50}=2^{100}\cdot3^{150}\)

\(73^{75}>72^{75}=\left(2^3\cdot3^2\right)^{75}=2^{225}\cdot3^{150}\)

mà \(2^{225}\cdot3^{150}>2^{100}\cdot3^{150}=108^{50}>107^{50}\)

nên \(73^{75}>107^{50}\)

d: \(2^{91}=\left(2^{13}\right)^7=8192^7\)

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

mà 8192>3125

nên \(2^{91}>5^{35}\)

e: \(A=72^{45}-72^{44}=72^{44}\left(72-1\right)=72^{44}\cdot71\)

\(B=72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)

mà 44>43

nên A>B

Bài 2:

a:

ĐKXĐ: x<>2023

\(\frac{x-2023}{4}=\frac{1}{x-2023}\)

=>\(\left(x-2023\right)\left(x-2023\right)=4\cdot1\)

=>\(\left(x-2023\right)^2=4\)

=>\(\left[\begin{array}{l}x-2023=2\\ x-2023=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+2023=2025\left(nhận\right)\\ x=-2+2023=2021\left(nhận\right)\end{array}\right.\)

b: \(\left(2x+1\right)^4=\left(2x+1\right)^6\)

=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)

=>\(\left(2x+1\right)^4\cdot\left\lbrack\left(2x+1\right)^2-1\right\rbrack=0\)

=>\(\left(2x+1\right)^4\cdot\left(2x+1-1\right)\left(2x+1+1\right)=0\)

=>\(2x\left(2x+1\right)^4\cdot\left(2x+2\right)=0\)

=>\(\left[\begin{array}{l}2x=0\\ 2x+1=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\\ x=-1\end{array}\right.\)

c: \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

=>\(\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

=>\(\left(3x-1\right)^{10}\cdot\left\lbrack\left(3x-1\right)^{10}-1\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(3x-1\right)^{10}=0\\ \left(3x-1\right)^{10}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-1=0\\ \left(3x-1\right)^{10}=1\end{array}\right.\)

=>\(\left[\begin{array}{l}3x-1=0\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=\frac23\\ x=0\end{array}\right.\)

d: Sửa đề \(2^{x+1}\cdot3^{y}=12^{x}\)

=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)

=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)

a)4^50=(2^2)^50=2^100

Vậy 2^100=4^50

b) 4^3x5^3=(4x5)^3=20^3

Vì 20^3>19^3 nên 4^3x5^3>19^3

Tìm x:

3^2x4^2:(x-2)=12

(3x4)^2:(x-2)=12

12^2:(x-2)-12

x-2=12^2:12

x-2=12

x=12+2

x=14 

a, 2¹⁰⁰ và 4⁵⁰

Ta có : 

4⁵⁰=(2²)⁵⁰=2¹⁰⁰

Vì 2¹⁰⁰ = 2¹⁰⁰

=> 4⁵⁰ = 2¹⁰⁰

KHÓ QUÁ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

bạn ơi sao bài này khó thế?

 Để tớ ghi đề giùm cho các bạn hiểu :

\(11^{21}+1\div11=121\)

\(4^{2x}+1=64\)

So sánh

\(10^{30}...2^{100}\)

\(2^{98}...9^{42}\)

bài 1

4^2x+1 = 64

=> 4^2x+1 = 4³

=> 2x + 1 = 3

=> 2x = 2

=> x = 1

bài 2

10³⁰ = ( 10³ )¹⁰ = 1000¹⁰

2¹⁰⁰ = ( 2¹⁰ )¹⁰ = 1024¹⁰

=> 1000¹⁰ < 1024¹⁰

=> 10³⁰ < 2¹⁰⁰

2⁹⁸ = ( 2⁷ )¹⁴ = 128¹⁴

9⁴² = ( 9³ )¹⁴ = 729¹⁴

=> 128¹⁴ < 729¹⁴

=> 2⁹⁸ < 9⁴²