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20 tháng 3 2017

Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)

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

\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)

\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)

Thay A vào biểu thức ta dc

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)

\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)

CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK

20 tháng 3 2017

sai rôi bạn ơi

1 tháng 8 2017

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)

\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)

\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)

\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)

1 tháng 8 2017

Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)

22 tháng 12 2016

sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)

30 tháng 7 2019

\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)

9 tháng 1 2020

\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)

=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)

=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)

=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0

=> x + 2017 = 0 

=> x = -2017

9 tháng 1 2020

\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)

\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)

\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

Dễ thấy cái ngoặc to < 0

=> x=-2017

20 tháng 3 2017

ko biết làm thì thôi