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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= \(\frac{5}{3}\) ( \(\frac{1}{4}\) - \(\frac{1}{28}\)) = \(\frac{5}{3}\) . \(\frac{3}{14}\)= \(\frac{5}{14}\)
Xét N ta có :
N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-15}{10^{2006}}\)
N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)+\(\frac{-8}{10^{2006}}\)
Xét M ta có :
M = \(\frac{-15}{10^{2005}}\)+\(\frac{-7}{10^{2006}}\)
M = \(\frac{-8}{10^{2005}}\)+\(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}\)< \(\frac{-8}{10^{2005}}\) => N < M
Xét A ta có
A=\(\frac{-7}{10^{2005}}\) + \(\frac{-15}{10^{2006}}\)
A=\(\frac{-7}{10^{2005}}\) +\(\frac{-8}{10^{2006}}\) +\(\frac{-7}{10^{2006}}\)
Xét B ta có
B=\(\frac{-15}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)
B=\(\frac{-8}{10^{2005}}\) + \(\frac{-7}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)
Vì \(\frac{-8}{10^{2006}}\) >\(\frac{-8}{10^{2005}}\) nên A>B
S=10/2.12+10/12.22+10/22.32+10/32.42+.......+10/2002.2012
S=1/2-1/12+1/12-1/22+1/22-1/32+1/32-1/42+.....+1/2002-1/2012
S=1/2-1/2012
S=????
bạn tự tính nhé
S=10.1/10{1/2-1/12+1/12-1/22+1/22-1/32+...+1/2002-1/2012}
=1/2-1/2012
=1005/2012
\(S = \frac{1}{3} +\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28} \)
\(S=\frac{1}{3}+\frac{1}{3}.\frac{1}{2}+\frac{1}{5}.\frac{1}{2}+\frac{1}{5}.\frac{1}{3}+\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{4} \)
\(S=\frac{1}{3}(1+\frac{1}{2})+\frac{1}{5}(\frac{1}{2}+\frac{1}{3})+\frac{1}{7}(\frac{1}{3}+\frac{1}{4})\)
\(S=\frac{1}{3}.\frac{3}{2}+\frac{1}{5}.\frac{5}{6}+\frac{1}{7}.\frac{7}{12}\)
\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\)
\(S=\frac{6}{12}+\frac{2}{12}+\frac{1}{12}\)
\(S=\frac{9}{12}\)
\(S=\frac{3}{4}\)
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
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!
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{14}\)