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17 tháng 9 2017

buom gai chau au

11 tháng 9 2016

\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\) 

\(\Leftrightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1=\frac{x+2}{2016}+1+\frac{x+1}{2017}+1\)

\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\) 

\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\) 

\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)  

Vì: \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\) 

\(\Rightarrow x+2018=0\Rightarrow x=-2018\)

5 tháng 3 2023

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

 

30 tháng 8 2017

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