Tìm x
X+1 trên 2017 cộng x+1 trên 2016. =x+1 trên 2015 cộng x+1 trên 2014
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\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+2}{2016}+\frac{x+1}{2017}\)
\(\Leftrightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1=\frac{x+2}{2016}+1+\frac{x+1}{2017}+1\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2016}+\frac{x+2018}{2017}\)
\(\Leftrightarrow\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2016}-\frac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
Vì: \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
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
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