\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)=x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{2016}>\frac{1}{2017}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}>\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\)
=> x = 0
Vậy x = 0
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=0\).Do \(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\ne0\)
Vậy x=0
