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\(\frac{2}{60}+\frac{2}{63}+\frac{2}{63}+\frac{2}{66}+....+\frac{2}{117}+\frac{2}{120}+\frac{2}{2003}\)
Ta có:
A = \(\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2016}\)
\(=2.\left(\frac{1}{60.63}+\frac{1}{63.66}+...+\frac{1}{117.120}\right)+\frac{2}{2016}\)
\(=2.\frac{1}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2016}\)
\(=\frac{2}{3}.\frac{1}{120}+\frac{2}{2016}\)
\(=\frac{1}{180}+\frac{2}{2016}\)
B = \(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2016}\)
\(=\frac{5}{4}.\frac{1}{80}+\frac{5}{2016}\)
\(=\frac{1}{64}+\frac{5}{2016}\)
Vì \(\frac{1}{64}>\frac{1}{180}\) và \(\frac{5}{2016}>\frac{2}{2016}\) nên B > A
Vậy B > A
Để mình làm lại nguyên bài cho dễ hiểu nhé
\(A=\frac{2}{60.63}+\frac{2}{63.66}+\frac{2}{66.69}+..+\frac{2}{117.120}+\frac{2}{2011}\)
\(=\frac{2}{3}\left(\frac{2}{60}-\frac{2}{63}+\frac{2}{63}-\frac{2}{66}+\frac{2}{66}-\frac{2}{69}+...+\frac{2}{117}-\frac{2}{120}\right)+\frac{2}{2011}\)
\(=\frac{2}{3}\left(\frac{2}{60}-\frac{2}{120}\right)+\frac{2}{2011}=\frac{2}{3}.\frac{1}{60}+\frac{2}{2011}=\frac{4382}{361980}\)
Sorry nhé! nãy giờ nhìn không kĩ đề
\(A=\frac{2}{60.63}+\frac{2}{63.66}+\frac{2}{66.69}+...+\frac{2}{117.120}\)
\(=\frac{2}{3}\left(\frac{2}{60}-\frac{2}{63}+\frac{2}{63}-\frac{2}{66}+\frac{2}{66}-\frac{2}{69}+...+\frac{2}{117}-\frac{2}{120}\right)\)
\(=\frac{2}{3}\left(\frac{2}{60}-\frac{2}{120}\right)=\frac{2}{3}.\frac{1}{60}=\frac{2}{180}\)
Suy ra \(A=\frac{2}{180}\)
Đề sai. Sửa đề: Ta có: \(B=\frac{5}{40.44}+\frac{5}{44.48}+\frac{5}{48.52}+...+\frac{5}{76.80}+\frac{5}{2011}\)
\(=\left(\frac{5}{40.44}+\frac{5}{44.48}+\frac{5}{48.52}+...+\frac{5}{76.80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40.44}+\frac{1}{44.48}+\frac{1}{48.52}+..+\frac{1}{76.80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+\frac{1}{48}-\frac{1}{52}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2011}=\frac{5}{4}.\frac{1}{80}+\frac{5}{2011}=\frac{1}{64}+\frac{5}{2011}\)
bạn CTV tth làm sai rồi nhé
\(\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-...-\frac{1}{76}+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2011}=\frac{5}{4}.\frac{1}{80}+\frac{5}{2011}\)
\(=\frac{1}{64}+\frac{5}{2001}\)