\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+....+\frac{5}{25.28}\)

\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{14}\)

H=\(\frac{5}{28}\) + \(\frac{5}{70}\) + \(\frac{5}{130}\) +...+ \(\frac{5}{700}\)

H= \(\frac{5}{4.7}\) + \(\frac{5}{7.10}\) + \(\frac{5}{10.13}\) +...+\(\frac{5}{25.28}\)

H= \(\frac{5}{3}\) (\(\frac{1}{4}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - \(\frac{1}{13}\) +...+ \(\frac{1}{25}\) - \(\frac{1}{28}\))

H= \(\frac{5}{3}\) (\(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) - \(\frac{1}{10}\)+...+ \(\frac{1}{25}\)- \(\frac{1}{25}\) - \(\frac{1}{28}\))

H= \(\frac{5}{3}\) ( \(\frac{1}{4}\) - \(\frac{1}{28}\)) = \(\frac{5}{3}\) . \(\frac{3}{14}\)= \(\frac{5}{14}\)

Ta có:  

\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

Ta lại có:

\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}\)

Hay A<B

A<B

ta co:B=20¹⁰-1/20¹⁰-3>1

=>B>20¹⁰-1+2/20¹⁰-3+2=20¹⁰+1/20¹⁰-1=A

vay A<B

B=\(\frac{2011^{10}-1}{2011^{10}-3}\) <1 => \(\frac{2011^{10}-1}{2011^{10}-3}\) < \(\frac{2011^{10}-1+2}{2011^{10}-3+2}\) = \(\frac{2011^{10}+1}{2011^{10}-1}\) = A

=> B<A

Cảm ơn bạn nhiều nha giải ra lại thấy dễ ak

\(A=\frac{10^8+2}{10^8-1}=\frac{\left(10^8-1\right)+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{\left(10^8-3\right)+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(1+\frac{3}{10^8-1}<1+\frac{3}{10^8-3}\) nên A < B

Ta có : 

 A = 10⁸ + 2 / 10 ⁸ - 1 = 1 + 3 / 10 ⁸ - 1

B = 10⁸ / 10 ⁸ - 3 =  1 + 3 / 10⁸ - 3

Vì 3/ 10⁸ - 1 < 3 / 10⁸ - 3=> A < B

Dễ thấy B < 1 vì 10²⁰¹¹ + 1 < 10²⁰¹² + 1. Áp dụng tính chất nếu \(\frac{a}{b}<1\) thì \(\frac{a}{b}<\frac{a+m}{b+m}\) ta có :

\(B=\frac{10^{2011}+1}{10^{2012}+1}<\frac{\left(10^{2011}+1\right)+9}{\left(10^{2012}+1\right)+9}=\frac{10^{2011}+10}{10^{2012}+10}=\frac{10.\left(10^{2010}+1\right)}{10.\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)

Vậy A > B

mình nghĩ là A<B

Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé! 

a) Ta có : 

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{151}=3^{150}\cdot3=\left(3^2\right)^{75}\cdot3=9^{75}\cdot3\)

Mà \(9^{75}>8^{75}=>9^{75}\cdot3>8^{75}=>3^{151}>2^{225}\)

b) Nhân cả vế A lẫn vế B với 10²⁰⁰⁵, ta có : 

\(10^{2005}A=-7+\frac{-15}{10}=\frac{-70}{10}+\frac{-15}{10}=\frac{-85}{10}\)

\(10^{2005}B=-15+\frac{-7}{10}=\frac{-150}{10}+\frac{-7}{10}=\frac{-157}{10}\)

Mà \(\frac{-85}{10}>\frac{-157}{10}=>10^{2005}A>10^{2005}B\)

\(=>A>B\)

Chúc bạn học tốt!