anh Tú à? chắc giờ này còn onl,nhưng tag kiểu đó chắc sáng mai đó bạn

Phải như vậy nè:

C1: Nguyễn Huy Tú

C2: @Nguyễn Huy Tú

a) 32⁵⁰ = 2²⁵⁰ < 2²⁵⁵ = 8⁸⁵

27⁵¹ = 3¹⁷¹ > 3¹⁷⁰ = 9⁸⁵

Vì 8⁸⁵ < 9⁸⁵ nên 32⁵⁰ < 27⁵¹

b) 31⁹ < 32⁹ = 2⁴⁵

9¹⁶ > 8¹⁶ = 2⁴⁸

Vì 2⁴⁵ < 2⁴⁸ nên 31⁹ < 9¹⁶

c) Chịu :)

a: B=31/39=62/78=1-16/78; A=35/37=280/296=1-16/296

16/78>16/296

=>B<A

b: C=23/27=46/54=1-8/54

D=1-8/29

8/29>8/54

=>D<C

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!!

câu d thì dễ

Thế giúp vs bn ơi

a) x lớn hơn 120

b) x=8

2) a) 2002/2001 lớn hơn

b) 2015/2018 lớn hơn

c) 27/37 lớn hơn

Đúng thì like giúp mik nha. Thx bạn

bạn có chs tiktok ko

A=1+2+2^2+2^3+....+2^9

2A=2+2^2+2^3+....+2^10

2A-A=2^10-1

A=2^10-1/2

B=5.2^8=(2^2+1).2^8=2^10+2^8

=>B>A

2A = 2(1 + 2 + 2² + .... + 2⁹ )

= 2 + 2² + 2³ + ..... + 2¹⁰

2A - A = (2 + 2² + 2³ + ..... + 2¹⁰) - (1 + 2 + 2² + .... + 2⁹ )

A = 2¹⁰ - 1  

B = 5.2⁸ = (2² + 1).2⁸ = 2¹⁰ + 2⁸

2¹⁰ - 1 < 2¹⁰ + 2⁸

=> A < B

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)