Ta có : +) A= 1/5 -1/7 +1/7 -1/12 +1/12 - 1/19 +1/19 - 1/28 +1/28 - 1/39 +1/30.40  ⇔ A=1/5 -1/39 +1/30.40

+) B= 2.(1/5.8 +1/8.11 +1/11.14 +1/14.17 + 1/17.20 ) 

⇔B=2. 1/3.(1/5 - 1/8 +1/8 - 1/11 +1/11- 1/14 +1/14 -1/17 +1/17 -1/20 )

⇔B=2/3.( 1/5-1/20 )  Ta luôn có :B luôn <1/5 - 1/20

Mà 1/5 -1/20 <1/5 -1/39 +1/30.40 =A 

⇒ A>B (dpcm) Tích mình với nha bn .

Đây nha bạn:

A=5.72​+7.125​+12.197​+19.289​+28.3911​+39.401​

$=\genfrac{}{}{0ex}{}{7-5}{5.7}+\genfrac{}{}{0ex}{}{12-7}{7.12}+\genfrac{}{}{0ex}{}{19-12}{12.19}+\genfrac{}{}{0ex}{}{28-19}{19.28}+\genfrac{}{}{0ex}{}{39-28}{28.39}+\genfrac{}{}{0ex}{}{40-39}{39.40}$

$=\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{7}+\genfrac{}{}{0ex}{}{1}{7}-\genfrac{}{}{0ex}{}{1}{12}+\genfrac{}{}{0ex}{}{1}{12}-\genfrac{}{}{0ex}{}{1}{19}+\genfrac{}{}{0ex}{}{1}{19}-\genfrac{}{}{0ex}{}{1}{28}+\genfrac{}{}{0ex}{}{1}{28}-\genfrac{}{}{0ex}{}{1}{39}+\genfrac{}{}{0ex}{}{1}{39}-\genfrac{}{}{0ex}{}{1}{40}$

=15−140=740=

\(A=\frac{2}{5.7}+\frac{5}{7.12}+\frac{7}{12.19}+\frac{9}{19.28}+\frac{11}{28.39}+\frac{1}{39.40}\)

\(=\frac{7-5}{5.7}+\frac{12-7}{7.12}+\frac{19-12}{12.19}+\frac{28-19}{19.28}+\frac{39-28}{28.39}+\frac{40-39}{39.40}\)

\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{19}+\frac{1}{19}-\frac{1}{28}+\frac{1}{28}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(=\frac{1}{5}-\frac{1}{40}=\frac{7}{40}\)

\(B=\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}\)

\(=\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}+\frac{2}{340}\)

\(=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}+\frac{2}{17.20}\)

\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{2}{3}\left(\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}+\frac{20-17}{17.20}\right)\)

\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{1}{10}\)

\(\frac{A}{B}=\frac{\frac{7}{40}}{\frac{1}{10}}=\frac{7}{4}\)

ta tách 2/5x7 = 2/5-2/7 tách những cái kia tương tự góp vào rồi tính

bn lm sai ùi

Sửa đề: 39*40

\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{39}-\dfrac{1}{40}=\dfrac{1}{5}-\dfrac{1}{40}=\dfrac{7}{40}\)

\(B=\dfrac{2}{3}\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{17\cdot20}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

=2/3*3/20=2/20=1/10=4/40<A

cacwj con

=307/990

Câu C giải rồi

\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)

\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)