Tính A:[1+1/1.3].[1+1/2.4]......[1+1/2013.2015]
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DN
1
NM
1
KB
21 tháng 5 2017
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)
=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)
=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)
=> đpcm
S
10 tháng 4 2018
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)
\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)
\(=\frac{2014\cdot2}{1\cdot2015}\)
\(=\frac{4028}{2015}\)
29 tháng 3 2023
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
TA
2
HP
12 tháng 5 2016
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
Vậy A=1007/2015
TN
12 tháng 5 2016
\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(2A=1-\frac{1}{2015}\)
\(A=\frac{2014}{2015}:2\)
\(A=\frac{1007}{2015}\)
12 tháng 5 2016
1/1.3+1/3.5+...+1/2013.2015
=1/2.(1/1-1/3+1/3-1/5+...+1/2013-1/2015)
=1/2.(1/1-1/2015)
=1/2.2014/2015
=1007/2015
12 tháng 5 2016
A=1/1.3+1/3.5+1/5.7+...+1/2013.2015
2A=2.(1/1.3+1/3.5+1/5.7+...+1/2013.2015)
=2/1.3+2/3.5+2/5.7+...+2/2013.2015
=1-1/3+1/5-1/7+1/7-1/9+...+1/2013-1/2015
=1-1/2015
=2014/2015
=>2A=2014/2015=>A=1007/2015
VM
1
HN
16 tháng 5 2021
\(B=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{2013\times2015}\\
2B=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{2013\times2015}\\
2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\\2B=1-\frac{1}{2015}=\frac{2014}{2015}\\
\Rightarrow B=\frac{2014}{2015}
\div2=\frac{1007}{2015}\)
21 tháng 5 2017
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}< \frac{3}{4}\)
\(\Leftrightarrow A=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\left(1-\frac{1}{2015}\right)+\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(\Leftrightarrow A=\frac{2014}{2015}+\frac{1007}{2016}\)
\(\Leftrightarrow A=1,5\)
Đổi \(\frac{3}{4}=0,75\)
Vì 0,75 < 1,5
Nên ko thể CM
21 tháng 5 2017
Bài này mà cũng hỏi thì đừng có thi nữa. đợi vài ngày sau có đáp án nhé.