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16 tháng 9 2021
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
NT
1
20 tháng 6 2017
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
TD
0
27 tháng 9 2019
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\Leftrightarrow\) \(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{x+4}{2015}+1+\frac{x+5}{2014}+1+\frac{x+6}{2013}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{x+2019}{2015}+\frac{x+2019}{2014}+\frac{x+2019}{2013}\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}-\frac{x+2019}{2015}-\frac{x+2019}{2014}-\frac{x+2019}{2013}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)\)\(=0\)
Lại có: \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\) \(\ne\) \(0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=0-2019=-2019\)
Vậy x= -2019
27 tháng 8 2021
\(x-\dfrac{1}{2017}+x-\dfrac{2}{2016}=x-\dfrac{3}{2015}+x-\dfrac{4}{2014}\)
\(\Rightarrow x+x-x-x=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(\Rightarrow0=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\) (vô lí)
27 tháng 8 2021
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\Leftrightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\Leftrightarrow x=2018\)
Ta có : \(\frac{x+1}{2013}+\frac{x+1}{2014}+\frac{x+1}{2015}=\frac{x+1}{2016}+\frac{x+1}{2017}\)
\(\Rightarrow\) \(\frac{x+1}{2013}+\frac{x+1}{2014}+\frac{x+1}{2015}-\frac{x+1}{2016}-\frac{x+1}{2017}=0\)
\(\Rightarrow\) \(\left(x+1\right)\left(\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
Vì \(\left(\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)\ne0\)
Nên : x + 1 = 0
Vậy x = -1