\(\left(1-\frac{52}{53}\right)+\left(\frac{105}{106}-1\right)+\left(\frac{158}{159}-1\right)=\frac{\left|x\right|}{318}\)
⇔\(\frac{1}{53}+\frac{-1}{106}+\frac{-1}{159}=\frac{\left|x\right|}{318}\)
⇔\(\frac{1}{138}=\frac{\left|x\right|}{318}\)
⇒\(\left|x\right|=1\)
⇔\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy x∈\(\left\{-1;1\right\}\)