\(\left(5^{2013}+5^{2015}\right):\left(5^{2011}+5^{2013}\right)\)
K
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13 tháng 5 2022
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
LB
20 tháng 4 2018
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
LB
20 tháng 4 2018
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
29 tháng 6 2021
Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
20 tháng 3 2020
a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)
\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)
Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015
\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)
<=>S=-1007+2015
<=> S=1008
27 tháng 7 2017
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
\(=\frac{3}{5}.\frac{5}{7}...\frac{2011}{2013}.\frac{2013}{2015}\)
\(=\frac{3.5...2011.2013}{5.7...2013.2015}\)
\(=\frac{3}{2015}\)
M
3 tháng 12 2017
\(=\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right).\frac{2012}{2013}\)
\(=\left(\frac{8}{4}-\frac{5}{5}\right).\frac{2012}{2013}\)
\(=\left(2-1\right).\frac{2012}{2013}\)
\(=\frac{2012}{2013}\)
3 tháng 12 2017
\(=\frac{2012}{2013}.\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{5}{4}+\frac{3}{4}-\frac{2}{5}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{8}{4}-\frac{5}{5}\right)=\frac{2012}{2013}.\left(1-1\right)=\frac{2012}{2013}.0\)