các bạn giúp mình với nhanh lên nhé.Mình sẽ cho

A>B

BẠN Ạ

Ta có: \(A=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\)

\(=1+\left(\frac12+\frac13+\frac16\right)+\left(\frac14+\frac15\right)+\left(\frac17+\frac18+\frac19\right)+\frac{1}{10}\)

\(=1+1+\frac{9}{20}+\frac{1}{10}+\left(\frac17+\frac18+\frac19\right)\)

=2,55+\(\left(\frac17+\frac18+\frac19\right)\) >2,3

=>A>B

\(A=\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}\)

\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)

\(\frac{-9}{10^{2010}}>\frac{-19}{10^{2010}}\)

\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-9}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)

\(\Rightarrow A>B\)

bn vào /h7.net/hoi-dap/toan-6/so-sanh-a-3-10-1-3-9-1-va-b-3-9-1-3-8-1--faq205231.html

Trả lời:

A = \(\frac{3^{10}+1}{3^9+1}=\frac{3.3^9+1}{3.3^8+1}=\frac{3^9+1}{3^8+1}\)= B

_Học tốt bạn nha_

\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)

\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

Có: \(\frac{1}{1+5+5^2+...+5^8}>0\)              và      \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)

\(\Rightarrow A>B\)

a)1024^9=(2^10)^9=2^90

2^100>2^90

b)125^80>20^118

c)9^12=(3^2)^12=3^24

27^10=(3^3)^10=3^30

3^24<3^30=9^12<27^10

tích nha

Ta có 10⁸ > 1 

       và 10⁸ < 10⁹

=) A > 1

Ta có : 10²¹ > 10²⁰

            10²¹ > 1 

=) B < 1

Vậy A > B