$a)\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}$

$=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}$

$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}$

$=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$

b) Đặt $B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{18.20}$

$=>\dfrac{1}{2}B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}$

$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}$

$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}$

$=>B=\dfrac{9}{10}$

c) $\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}$

$=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}$

$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}$

$=1-\dfrac{1}{9}=\dfrac{8}{9}$

d) Viết lại đề rõ ràng nha bạn.

Đặt \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)

\(A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{18-16}{16.18}\)

\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)

\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{18}\right)\)

\(A=\frac{4}{2}.\frac{4}{9}\)

\(\Rightarrow A=\frac{8}{9}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)

\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{18}\right)\)

\(=2.\frac{4}{9}\)

\(=\frac{8}{9}\)

I.\(B=9,8+8,7+7,6+...+2,1-1,2-2,3-3,4-...-8,9\)

\(B=\left(9,8-8,9\right)+\left(8,7-7,8\right)+\left(7,6-6,7\right)+...+\left(2,1-1,2\right)\)

\(B=0,9+0,9+0,9+...+0,9\) ( 8 số 0,9 )

\(B=7,2\)

II.

\(\left(a\right)\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{19\cdot20}\)

\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{19}-\frac{1}{20}\right)\)

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

\(=2\cdot\frac{19}{20}=\frac{19}{10}\)

\(\left(b\right)\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{17\cdot19}+\frac{4}{19\cdot21}\)

\(=2\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\left(1-\frac{1}{21}\right)\)

\(=2\cdot\frac{20}{21}=\frac{40}{21}\)

\(\left(c\right)\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{16\cdot18}+\frac{4}{18\cdot20}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

cảm ơn  bạn

\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{16\cdot18}+\dfrac{4}{18\cdot20}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{16\cdot18}+\dfrac{2}{18\cdot20}\right)\)

\(=2\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)

\(=2\left(1-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{20}{20}-\dfrac{1}{20}\right)\)

\(=2\cdot\dfrac{19}{20}\)

\(=\dfrac{19}{10}\)

\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+.....+\dfrac{4}{16.18}+\dfrac{4}{18.20}\)

\(=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{16.18}+\dfrac{1}{18.20}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)\(=2\left(1-\dfrac{1}{20}\right)=2.\dfrac{19}{20}=\dfrac{19}{10}\)

\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+..................+\dfrac{4}{16.18}+\dfrac{4}{18.20}\)

\(=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{16.18}+\dfrac{2}{18.20}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...........+\dfrac{1}{18}-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)

\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)

1/ 2.4 + 1/4.6 + ...+1/18.20

= 1/2 - 1/4 + 1/4 -1/6 + .... + 1/18.20

trừ hết đi cho nhau cuối cùng:

= 1/2 - 1/20 = 9/20

a)   \(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}=\frac{1}{5}\)

b)   \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}\)

\(=1-\frac{1}{17}=\frac{16}{17}\)

hok tốt ...

a)\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\frac{2}{8\cdot10}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(A=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\)

b)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)

a) 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10

= 2/2.2.4 + 2/2.4.6 + 2/2.6.8 + 2/2.8.10 ( nhân cả tử và mẫu với 2)

= 1/2 .( 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 )

= 1/2 .(1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/10)

= 1/2.(1/2 - 1/10)

= 1/2.( 5/10 - 1/10) = 1/2.4/10 = 2/10 = 1/5

b) 4/1.5+ 4/5.9 + 4/ 9.13 + 4/13.17

= 1- 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17

= 1- 1/17

= 16/17 

\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)

\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)