giúp mik ạ

Sửa đề: bỏ 1/30*40

\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{19}+...+\dfrac{1}{28}-\dfrac{1}{39}\)

=1/5-1/39

=34/195

Còn B là biểu thức nào vậy bạn?

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 .

A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\)  + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\) 

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

B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))

B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))

B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)

B = \(\dfrac{1}{10}\)  = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)

Vậy A > B 

giúp với

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

Đây nha bạn:

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

=7−55.7+12−77.12+19−1212.19+28−1919.28+39−2828.39+40−3939.40=5.77−5​+7.1212−7​+12.1919−12​+19.2828−19​+28.3939−28​+39.4040−39​

=15−17+17−112+112−119+119−128+128−139+139−140=51​−71​+71​−121​+121​−191​+191​−281​+281​−391​+391​−401​

=15−140=740=

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

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

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

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

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

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

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

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

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

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

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

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

=307/990