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\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)
\(\Leftrightarrow x=2014\)
Vậy \(x=2014\)
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)
Vậy x = 2014
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
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 \(\left(S-P\right)^{2016}=0\)
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 \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)
\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)
\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)
\(\frac{x-2015}{2014}+\frac{x-2015}{2013}-\frac{x-2015}{2012}-\frac{x-2015}{2011}=0\)
\(\left(x-2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)
\(\Rightarrow x-2015=0\)
\(x=0+2015\)
\(x=2015\)
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
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+4}{2012}+1+\dfrac{x+3}{2013}+1=\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)
\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\left(\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\right)=0\)
\(\Leftrightarrow x+2016.\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
Vì \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
Vậy \(x=-2016\) tại biểu thức \(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)
Theo đề ta có: x+4/2012+x+3/2013=x+2/2014+x+1/2015
=>x+4/2012+x+3/2013-x+2/2014+x+1/2015=0
=>( x+4/2012+1)+(x+3/2013+1)-(x+2/2014+1)+(x+1/2015+1)
=>x+2016/2012+x+2016/2013-x+2016/2014-x+2016/2015=0
=>x+2016.(1/2012+1/2013-1/2014-1/2015)=0
Do 1/2012+1/2013-1/2014-1/2015>0
nên x+2016=0
=>x=-2016
Vậy x=-2016
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2013.2014}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}=1-\dfrac{1}{2014}=\dfrac{2013}{2014}\left(đpcm\right)\)