CM : (1 - 1/2) (1-1/3) ... (1- 1/2015) < 1/2014
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12 tháng 3 2019
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
TT
0
SN
0
NM
0
27 tháng 7 2017
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
S
1
S
1 tháng 5 2017
\(A=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right)......\left(1-\frac{2015}{2014}\right)\)
\(=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right).....\left(1-\frac{2014}{2014}\right)\left(1-\frac{2015}{2014}\right)\)
\(=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right)......0.\left(1-\frac{2015}{2014}\right)\)
\(=0\)
ND
0
SL
1
1 tháng 2 2022
\(B=\left(\dfrac{1}{2015}+1\right)+\left(\dfrac{2}{2014}+1\right)+\left(\dfrac{3}{2013}+1\right)+...+\left(\dfrac{2014}{2}+1\right)+1\)
\(=\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2016}\)
=>B:A=2016
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2014}{2015}\)
\(=\frac{1}{2015}< \frac{1}{2014}\).