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\(A=\frac{1}{1.4}+\frac{1}{2.7}+...+\frac{1}{67.70}\)
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{67.70}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{70}\)
\(3A=1-\frac{1}{70}=\frac{69}{70}\)
\(A=\frac{69}{70}:3=\frac{23}{70}\)
vì \(\frac{23}{70}< 1\)
nên \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{67.70}< 1\)
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(A=1-\frac{1}{16}=\frac{15}{16}\)
Ta có :
\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)
\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(\frac{3}{5}B=1-\frac{1}{16}\)
\(B=\frac{15}{16}:\frac{3}{5}\)
\(B=\frac{25}{16}\)
Ủng hộ mk nha !!! ^_^
\(B=\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+\frac{5}{13.16}\)
\(\frac{3}{5}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(\frac{3}{5}B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(\frac{3}{5}B=1-\frac{1}{16}\)
\(B=\frac{15}{16}:\frac{3}{5}\)
\(B=\frac{25}{16}\)
\(^{\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{10}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(1-\frac{1}{46}=\frac{45}{46}\)
Vì \(1-\frac{1}{46}< 1\)nên \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}< 1\)
Chúc bạn học tốt
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}\)
\(S=\frac{42}{43}< 1\)
#)Giải :
\(\frac{91}{1.4}+\frac{91}{4.7}+\frac{91}{7.11}+...+\frac{91}{88.91}\)
\(=\frac{91}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{88.91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{88}-\frac{1}{91}\right)\)
\(=\frac{91}{3}\left(1-\frac{1}{91}\right)\)
\(=\frac{91}{3}.\frac{90}{91}=30\left(đpcm\right)\)
#~Will~be~Pens~#
\(\frac{91}{1\cdot4}+\frac{91}{4\cdot7}+...+\frac{91}{88\cdot91}=\frac{1}{3}\left(91-\frac{91}{4}+\frac{91}{4}-\frac{91}{7}+...-\frac{91}{91}\right)\)
\(=\frac{1}{3}\left(91-1\right)=\frac{1}{3}\cdot90=30\)
A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)
A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{100}\)
A : 3 = \(\frac{99}{100}\)
A = \(\frac{297}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
\(B=\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
\(=\frac{1}{3022}\left(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+...+\frac{3022}{2014.1008}\right)\)
\(=\frac{1}{3022}\left(\frac{1008}{1008.2014}+\frac{2014}{1008.2014}+...+\frac{2014}{1008.2014}+\frac{1008}{1008.2014}\right)\)
\(=\frac{1}{3022}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+...+\frac{1}{2014}+\frac{1}{1008}\right)\)
\(=\frac{2}{3022}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
=> \(\frac{A}{B}=\frac{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
Vậy....
\(A=\frac{2}{3}\left[\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right]\)
\(A=\frac{2}{3}\left[\left[\frac{1}{1}-\frac{1}{4}\right]+\left[\frac{1}{4}-\frac{1}{7}\right]+...+\left[\frac{1}{97}-\frac{1}{100}\right]\right]\)
\(A=\frac{2}{3}\left[\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right]\)
\(A=\frac{2}{3}\left[1-\frac{1}{100}\right]=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
AI THẤY ĐÚNG ỦNG HỘ MIK NHÉ
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{2010.2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2010}-\frac{1}{2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}.\frac{2012}{2013}