B = 4/2.4 + 4/4.6 +4/6.8 + 4/8.10 +... +4/16.18 + 4/18.20 = ?
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13 tháng 8 2017
$a)\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}$
$=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}$
$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}$
$=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$
13 tháng 8 2017
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.
DP
25 tháng 6 2017
Đặt \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)
\(A=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{18-16}{16.18}\)
\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(A=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{18}\right)\)
\(A=\frac{4}{2}.\frac{4}{9}\)
\(\Rightarrow A=\frac{8}{9}\)
ML
25 tháng 6 2017
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{16.18}\)
\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{18}\right)\)
\(=2.\frac{4}{9}\)
\(=\frac{8}{9}\)
5 tháng 8 2020
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.
\(\left(a\right)\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{19\cdot20}\)
\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)\)
\(=2\cdot\frac{19}{20}=\frac{19}{10}\)
\(\left(b\right)\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{17\cdot19}+\frac{4}{19\cdot21}\)
\(=2\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\)
\(=2\left(1-\frac{1}{21}\right)\)
\(=2\cdot\frac{20}{21}=\frac{40}{21}\)
\(\left(c\right)\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{16\cdot18}+\frac{4}{18\cdot20}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
25 tháng 6 2017
\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{16\cdot18}+\dfrac{4}{18\cdot20}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{16\cdot18}+\dfrac{2}{18\cdot20}\right)\)
\(=2\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)
\(=2\left(1-\dfrac{1}{20}\right)\)
\(=2\left(\dfrac{20}{20}-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)
25 tháng 6 2017
\(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+.....+\dfrac{4}{16.18}+\dfrac{4}{18.20}\)
\(=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{16.18}+\dfrac{1}{18.20}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{16}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{20}\right)\)\(=2\left(1-\dfrac{1}{20}\right)=2.\dfrac{19}{20}=\dfrac{19}{10}\)
NT
1
27 tháng 5 2016
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
TH
3
N
29 tháng 7 2018
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 ...
BT
29 tháng 7 2018
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)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)
TM
25 tháng 4 2016
\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)
NT
1
29 tháng 4 2021
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
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