\(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\)

= \(4.\left(\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{23.27}\right)\)

=\(1.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{23.27}\right)\)

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

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

=\(1.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\)

= \(1.\dfrac{8}{27}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

a)1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72+89/90

=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90

=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)

=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]

=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)

=9 – (1 – 1/10) = 9 – 9/10 = 81/10

b)4/3.7 + 4/7.11 + 4/11.15 + 4/15.19 + 4/19.23 + 4/23.27

=4.(4/3.7 + 4/7.11 + ........+ 4/23.27 )

=1.( 1/3.7 + 1/7.11 + ......+ 1/23.27 )

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

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

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

=1.8/27

=8/27

c)1/10+1/40+1/88+1/154+1/138+1/340

=1/2.5 + 1/5.8 + 1/11.8 + 1/11.14 + 1/14.17 + 1/17.20

=1/3. (3/2.5  + 3/5.8 + 3/8.11 + 3/11.14 + 3/14.17 + 3/17.20 )

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

=1/3. ( 1/2 - 1/20 )

=1/3. 9/20

=3/20

P/S: CHÚC HOK TỐT !

4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )

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

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

4.\(\frac{12}{37}\)

48/37

Chắc là đề thiếu: \(y=\frac{1}{2}-\frac{1}{3\cdot7}-\frac{1}{7\cdot11}-\frac{1}{11\cdot15}-\frac{1}{15\cdot19}-\frac{1}{19\cdot23}-\frac{1}{23\cdot27}\)

\(y=\frac{1}{2}-\left(\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+...+\frac{1}{23\cdot27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{23\cdot27}\right)\)

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

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

Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\) 

             \(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)

             ..........................

             \(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)

=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)

=>                                                       =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)

Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)

\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{1023.1027}\)

\(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{1027}\)

\(=\frac{1024}{3081}\)

ai k mik mik k lại nha

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:

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

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

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

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

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

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

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

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

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