TÍNH A SẼ BÍT

TA CÓ:A=2/3x3/60x63+2/3x3/63x66+2/3x3/66x69+.....+2/3x3/117x120

           A=2/3x(3/60x63+3/63x66+3/66x69+.....+3/117x120

           A=2/3x(3/60-3/63+3/63-3/66+3/66-3/69+....+3/117-3/120)

           A=2/3x(3/60-3/120)

           A=2/3x1/40

           A=2/120=1/60

           1/60 VA 2/2011

           1/60 >2/2011

khó nhỉ

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 \(A=\frac{2}{180}\)

Ta có: \(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+\cdots+\frac{2}{117\cdot120}+\frac{2}{2003}\)

\(=\frac23\left(\frac{3}{60\cdot63}+\frac{3}{63\cdot66}+\cdots+\frac{3}{117\cdot120}\right)+\frac{2}{2003}\)

\(=\frac23\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+\cdots+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)

\(=\frac23\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}=\frac23\cdot\frac{1}{120}+\frac{2}{2003}=\frac{1}{60\cdot3}+\frac{2}{2003}=2\left(\frac{1}{360}+\frac{1}{2003}\right)\)

TA có: \(B=\frac{5}{40\cdot44}+\frac{5}{44\cdot48}+\cdots+\frac{5}{76\cdot80}+\frac{5}{2003}\)

\(=\frac54\left(\frac{4}{40\cdot44}+\frac{4}{44\cdot48}+\cdots+\frac{4}{76\cdot80}\right)+\frac{5}{2003}\)

\(=\frac54\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+\cdots+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)

\(=\frac54\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}=\frac54\cdot\frac{1}{80}+\frac{5}{2003}=5\left(\frac{1}{320}+\frac{1}{2003}\right)\)

\(A-B=\frac{2}{360}+\frac{2}{2003}-\frac{5}{320}-\frac{5}{2003}\)

\(=\frac{2}{360}-\frac{5}{320}-\frac{3}{2003}=\frac{2\cdot8-5\cdot9}{2880}-\frac{3}{2003}=\frac{-29}{2880}-\frac{3}{2003}<0\)

=>A<B

Bài này cũng dễ mà! Hãy cố gắng lên bạn sẽ làm được mà.

Con Ly Anh Tho kia biết trả lời thì trả lời đi chứ nói như vậy ai mà biết làm.