\(Cho\) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-...+\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2015}\)
\(B=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2014}+\dfrac{1}{2015}\)
Tính (S - B)2016
K
Khách
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3 tháng 4 2017
Ta có:
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(1+\dfrac{1}{2}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Mà \(P=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=P\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2016}=0^{2016}=0\)
Vậy \(\left(S-P\right)^{2016}=0\)
20 tháng 7 2017
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
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
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
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
NV
2
1 tháng 3 2017
\(\Leftrightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x}{1008}+\dfrac{x+3}{2013}+1\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x}{1008}+\dfrac{x+2016}{2013}\)
\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x}{1008}-\dfrac{x+2016}{2013}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(-\dfrac{x}{1008}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}\right)=0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
KT
1 tháng 3 2017
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x}{1008}+1+\frac{x+3}{2013}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+1008}{1008}+1+\frac{x+3}{2013}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{1008}+\frac{x+2016}{2013}\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}-\frac{1}{2013}\right)=0\)
vì \(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}+\frac{1}{2013}\right)\ne0\)nên
x+2016=0\(\Leftrightarrow\)x=-2016
Ta có:
*) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(\Rightarrow S=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Vậy \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0\)