\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)
<=>\(\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)=0\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
<=>\(\left|x\right|.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)
Vì \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow\left|x\right|=0\Rightarrow x=0\)
Vậy x=0
\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)
\(\Rightarrow\left|x.\left(\frac{1}{2015}+\frac{1}{2016}\right)\right|=\left|x.\left(\frac{1}{2016}+\frac{1}{2017}\right)\right|\)
\(\Rightarrow\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)
Mà \(\frac{1}{2015}+\frac{1}{2016}>\frac{1}{2016}+\frac{1}{2017}\)
=> |x| = 0
=> x = 0
Vậy x = 0