Tính B = 2014/1 + 2013/2 + 2013/3 + ... + 1/2014
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NT
1
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
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
BP
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
biết vậy mà vẫn đòi lấy ảnh! ok!
S
14 tháng 2 2020
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
\(B=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+....+\frac{1}{2014}\)
\(=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(1+\frac{1}{2014}\right)+1\)
\(=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(B=\frac{2014}{1}+\frac{2013}{2}+......+\frac{1}{2014}\)
\(B=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)+1\)
\(B=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(B=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)