Bài 1: Thực hiện phép tính:
A= (1+\(\frac{1}{1.3}\)).( 1+ \(\frac{1}{2.4}\)).( 1+\(\frac{1}{3.5}\))....(\(1+\frac{1}{2017.2019}\))
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TP
8 tháng 8 2017
Ta có:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}=\frac{1}{2}.\left(\frac{8}{9}+\frac{2}{5}\right)=\frac{1}{2}.\frac{58}{45}=\frac{29}{45}\)
LT
0
TD
3
XO
3 tháng 4 2020
Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)
3 tháng 4 2020
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)
=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)
=\(\frac{1}{2}.\frac{20}{21}\)
=\(\frac{20}{42}=\frac{10}{21}\)
18 tháng 12 2019
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{2018}{2019}\)
\(=\frac{2018}{4038}\)
\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính )
HN
18 tháng 12 2019
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)
\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow M=1-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2018}{2019}\)
Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)
\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)
Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)
Vậy M < \(\frac{1}{2}\)
3 tháng 3 2016
= 1/2. ( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +........+ 1/2013 - 1/2015)
= 1/2 . ( 1- 1/2015)
= 1007/2015
10 tháng 3 2016
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
10 tháng 3 2016
2*(1/1*3+1/3*5+.......+1/99*100)
=2*(2/1*3+2/3*5+.....+2/99*100)*1/2
=1/3-1/5+1/5-1/7+....+1/99-1/100
=1/3-1/100
=100/300-3/300
=97/300
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2017.2019}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2018^2}{2017.2019}\)
\(=\frac{2}{1}.\frac{2018}{2019}=\frac{4036}{2019}\)
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