Tìm x
(x+4/2014)+(x+3/2015)=(x+2/2016)+(x+1/2017)
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\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
\(x-\dfrac{1}{2017}+x-\dfrac{2}{2016}=x-\dfrac{3}{2015}+x-\dfrac{4}{2014}\)
\(\Rightarrow x+x-x-x=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(\Rightarrow0=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\) (vô lí)
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\Leftrightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\Leftrightarrow x=2018\)
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}+\frac{x+2018}{2017}=0\)
\(x+2018.\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\)\(\frac{x+1}{2017}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+2}{2016}+1\right)+\left(\frac{x+1}{2017}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(M\text{à:}\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
Ta có : \(\frac{x+5}{2012}+\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}+\frac{x}{2017}\)
\(\Rightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1+\frac{x+3}{2014}=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1+\frac{x}{2017}+1\)
\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}+\frac{x+2017}{2017}\)
\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}-\frac{x+2017}{2017}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\text{Mà
}\)\(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)\ne0\)
\(\text{Nên : }\) x + 2017 = 0
=> x = -2017
\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)
\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)
\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
\(\Rightarrow x+4031=0\)
\(\Rightarrow x=-4031\)
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)
\(\Leftrightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1=\frac{x+2}{2016}+1+\frac{x+1}{2017}+1\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
Vì: \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)