\(\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+...+\dfrac{4017}{2008.2009.2010}\) Rút gọn rồi tính
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ZD
0
ZD
0
TD
0
TM
16 tháng 3 2021
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
S
1
NA
Nguyễn An Ninh
VIP
14 tháng 5 2023
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{37.38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1482}\right)\)
\(A=\dfrac{1}{2}.\dfrac{370}{741}=\dfrac{185}{741}\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
\(\dfrac{3}{12}+\dfrac{1}{4}=\dfrac{3:3}{12:3}+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\dfrac{4}{10}+\dfrac{3}{5}=\dfrac{4:2}{10:2}+\dfrac{3}{5}=\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\)
\(\dfrac{12}{27}+\dfrac{2}{9}=\dfrac{12:3}{27:3}+\dfrac{2}{9}=\dfrac{4}{9}+\dfrac{2}{9}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{7}{3}+\dfrac{20}{15}=\dfrac{7}{3}+\dfrac{20:5}{15:5}=\dfrac{7}{3}+\dfrac{4}{3}=\dfrac{11}{3}\)
1
OY
26 tháng 9 2021
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
F
3
17 tháng 5 2022
\(2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{98.99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{50.99-1}{100.99}=\dfrac{4949}{9900}\)
\(S=\dfrac{4}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{6}{2.3.4}-\dfrac{1}{2.3.4}+...+\dfrac{4018}{2008.2009.2010}-\dfrac{1}{2008.2009.2010}\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2008.2010}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\right)\)
\(=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)-\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2008.2009.2010}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)
\(=\left(1-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2009.2010}\right)\)
\(=1-\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2009.2010}\)
\(=\dfrac{1}{2010}\left(\dfrac{1}{2009}-1\right)-\left(\dfrac{1}{2009}-1\right)\)
\(=\left(\dfrac{1}{2010}-1\right)\left(\dfrac{1}{2009}-1\right)=\dfrac{2009}{2010}.\dfrac{2008}{2009}=\dfrac{1004}{1005}\)