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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= \(\frac{5}{3}\) ( \(\frac{1}{4}\) - \(\frac{1}{28}\)) = \(\frac{5}{3}\) . \(\frac{3}{14}\)= \(\frac{5}{14}\)
Ta có:
\(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2B-B=B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Vậy \(A=\frac{1}{2^{10}}\)
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
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
a) Ta có: \(\frac{-9}{80}=\frac{\left(-9\right)x4}{80x4}=\frac{-36}{320}\) và \(\frac{17}{320}\)
b) Ta có: \(\frac{-7}{10}=\frac{\left(-7\right)x33}{10x33}=\frac{-231}{330}\) và \(\frac{1}{33}=\frac{1x10}{33x10}=\frac{10}{330}\)
c) Ta có:
\(\frac{-5}{14}=\frac{\left(-5\right)x10}{14x10}=\frac{-50}{140}\)
\(\frac{3}{20}=\frac{3x7}{20x7}=\frac{21}{140}\)
\(\frac{9}{70}=\frac{9x2}{70x2}=\frac{18}{140}\)
d) Ta có:
\(\frac{10}{42}=\frac{10x22}{42x22}=\frac{220}{924}\)
\(\frac{-3}{28}=\frac{\left(-3\right)x33}{28x33}=\frac{-99}{924}\)
\(\frac{-55}{132}=\frac{\left(-55\right)x7}{132x7}=\frac{-385}{924}\)
\(A=\frac{10^8+2}{10^8-1}=\frac{\left(10^8-1\right)+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{\left(10^8-3\right)+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
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
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
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!
\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+....+\frac{5}{25.28}\)
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{14}\)