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14 tháng 3 2017

\(B=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+....+\frac{1}{2014}\)

\(=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(1+\frac{1}{2014}\right)+1\)

\(=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)

\(=2015\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}+\frac{1}{2015}\right)\)

\(B=\frac{2014}{1}+\frac{2013}{2}+......+\frac{1}{2014}\)

\(B=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)+1\)

\(B=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)

\(B=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)

30 tháng 6 2018

ta có: \(A=\frac{2014^{2013}+1}{2014^{2013}-1}=\frac{2014^{2013}-1+2}{2014^{2013}-1}=1+\frac{2}{2014^{2013}-1}\)

\(B=\frac{2014^{2013}-1}{2014^{2013}-3}=\frac{2014^{2013}-3+2}{2014^{2013}-3}=1+\frac{2}{2014^{2013}-3}\)

\(\Rightarrow\frac{2}{2014^{2013}-1}< \frac{2}{2014^{2013}-3}\)

\(\Rightarrow1+\frac{2}{2014^{2013}-1}< 1+\frac{2}{2014^{2013}-3}\)

=> A < B

12 tháng 3 2019

\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)

=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)

11 tháng 2 2022

Ai trả lời đi please

30 tháng 8 2023

A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)

= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B

\(\Rightarrow\) \(\dfrac{A}{B}\)=2015

14 tháng 2 2020

\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)

\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)

14 tháng 2 2020

\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)

\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)

4 tháng 11 2018

Ta có: \(2014S=2014\left(1+2014+2014^2+2014^3+...+2014^{2013}\right)\)

\(2014S=2014+2014^2+2014^3+2014^4+...+2014^{2014}\)

\(2014S-S=\left(2014+2014^2+2014^3+2014^4+...+2014^{2014}\right)-\left(1+2014+2014^2+2014^3+...+2014^{2013}\right)\)

\(2013S=2014^{2014}-1\)

\(S=\dfrac{2014^{2014}-1}{2013}\)

\(P-S=\dfrac{2014^{2014}}{2013}-\dfrac{2014^{2014}-1}{2013}=\dfrac{1}{2013}\)

4 tháng 11 2018

cảm ơn bạn

\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)

\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)

\(=0\)

27 tháng 2 2022

bạn làm thế là sai rồi 

có 3 con 2014 cơ mà