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KC
0
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
PT
0
5 tháng 1 2016
Bài này nếu dùng hằng đẳng thức lớp 8 thì hay hơn.Thôi cứ làm vầy nhé:
1+1/1.3=2^2/1.3 ; 1+1/2.4=3^2/2.4 ; 1+1/3.5=4^2/3.5 ; ......;1+1/2007.2009=2008^2/2007.2009 Thấy quy luật rồi chứ!
ta được A=(2^2.3^2.4^2.....2008^2)/1.3.2.4.3.5.4.6.5.7...2007.2009 Chú ý quan sát và sắp xếp
để giản ước hết.
=(2^2.3^2.4^2....2008^2)/(1.2.3.4.5.6....2007.3.4.5.6....2007.2008.2009) Chuẩn đó
=(2^2.3^2.4^2....2008^2)/(2.3^2.4^2.5^2....2007^2.2008.2009) Viết ra nháp thì dễ nhìn hơn
=2^2.2008^2/2.2008.2009=2.2008/2009 Tựbấm máy và nhớ TICK đó.
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
N
1
30 tháng 10 2018
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....+\left(1+\frac{1}{99.101}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(=\frac{2.3.4.....100}{1.2.3.....99}.\frac{2.3.4.....100}{3.4.5.....101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
MH
17 tháng 10 2023
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
17 tháng 10 2023
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
DT
Tính S = 1.3/3.5 + 2.4/5.7 + 3.5/7.9 + ... + ( n-1)( n+1) / (2n-1)(2n+1) + ... + 1002.1004/2005.2007
2
KN
13 tháng 8 2020
\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)
\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)
\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)
\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)
\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)
\(\Rightarrow S=125,4372197\)
\(\)
PA
3 tháng 12 2016
\(A=\frac{1}{1\times3}+\frac{1}{2\times4}+\frac{1}{3\times5}+\frac{1}{4\times6}+\frac{1}{5\times7}+\frac{1}{6\times8}+\frac{1}{7\times9}+\frac{1}{8\times10}\)
\(2A=\frac{2}{1\times3}+\frac{2}{2\times4}+\frac{2}{3\times5}+\frac{2}{4\times6}+\frac{2}{5\times7}+\frac{2}{6\times8}+\frac{2}{7\times9}+\frac{2}{8\times10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\frac{1}{5}-\frac{1}{7}+\frac{1}{6}-\frac{1}{8}+\frac{1}{7}-\frac{1}{9}+\frac{1}{8}-\frac{1}{10}\)
\(2A=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\)
\(2A=\frac{58}{45}\)
\(A=\frac{58}{45}\div2\)
\(A=\frac{29}{45}\)
LT
3 tháng 12 2016
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{10}\)
\(=1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}=\frac{58}{45}\)
\(A=\frac{29}{45}\)