Ta có : \(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)
\(\Leftrightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)\)
Vì : \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)\ne0\)
Nên x = 0
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)=x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
\(\Rightarrow x.\left[\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\right]=0\)
\(\Rightarrow x=0\)\(\left(vi\left[\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\right]\right)\ne0\)
