\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

\(A=\frac{1}{2}-\frac{1}{3\times7}-\frac{1}{7\times11}-\frac{1}{11\times15}-...-\frac{1}{19\times23}-\frac{1}{23\times27}\)

\(=\frac{1}{2}-4\times\left(\frac{4}{3\times7}+\frac{4}{7\times11}+\frac{4}{11\times15}+...+\frac{4}{19\times23}+\frac{4}{23\times27}\right)\)

\(=\frac{1}{2}-4\times\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-4\times\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-4\times\frac{8}{27}\)

\(=\frac{1}{2}-\frac{32}{27}\)

\(=-\frac{37}{54}\)

A= 1/2- 1/4*[ 4/3*7 +4/7*11+ 4/11*15+...+4/19*23+ 4/23*27]

= 1/2- 1/4*[ 1/3- 1/7+ 1/7- 1/11+ 1/11- 1/15+ ...+ 1/19- 1/23+ 1/23- 1/27]

=1/2- 1/4*[1/3- 1/27]

=1/2- 1/4*8/27

=1/2- 2/27

=23/54

Ta có : \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}=\frac{1}{2}-\frac{2}{27}=\frac{23}{54}\)

Trả lời:

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

Học tốt 

a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)

a/b= 7/6 + 7/10 + 7/12

a/b= 7(1/6+1/10+1/12)

Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)

Bạn ơi cho mình hỏi dpcm là gì vậy?

dpcm: điều phải chứng minh

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

    \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

   \(=\frac{1}{2}-\frac{1}{50}\)

   \(=\frac{12}{25}\)

\(B=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{23.27}\)

    \(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\right)\)

    \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\right)\)

  \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

  \(=\frac{1}{4}.\frac{8}{27}=\frac{2}{27}\)

A = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/49.50 

A = 1/2 - 1/3 + 1/3  - 1/4 + 1/4 - 1/5 + ... + 1/49 - 1/50 

A = 1/2 - 1/50

A = 25/50 - 1/50

A = 24/50

A = 12/25

B = 1/3.7 + 1/7.11 + 1/11.15 + ... + 1/23.27

B = 1/4 . ( 4/3.7 + 4/7.11 + 4/11.15 + ... + 4/23.27 )

B = 1/4 . ( 1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/23 - 1/27 ) 

B = 1/4 . ( 1/3 - 1/27 )

B = 1/4 . ( 9/27 - 1/27 )

B = 1/4 . 8/27 

B = 2/27 

NHAN A VOI 3 RUI TU TINH

DỄ MÀ

4A=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

4A=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

4A=\(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

A=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)

\(4A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(4A=\frac{1}{3}-\frac{1}{111}\)

\(4A=\frac{36}{111}\)

\(A=\frac{36}{111}\div4\)

\(A=\frac{9}{111}\)