\(\dfrac{x}{186}=\left(1-\dfrac{3030}{3131}\right)+\left(\dfrac{6161}{6262}-1\right)+\left(\dfrac{929292}{939393}-1\right)\\ \dfrac{x}{186}=\left(1-\dfrac{30}{31}\right)+\left(\dfrac{61}{62}-1\right)+\left(\dfrac{92}{93}-1\right)\\ \dfrac{x}{186}=\dfrac{1}{31}+\dfrac{-1}{62}+\dfrac{-1}{93}\\ \dfrac{x}{186}=\dfrac{6}{186}+\dfrac{-3}{186}+\dfrac{-2}{186}\\ \dfrac{x}{186}=\dfrac{1}{186}\\ \Rightarrow x=1\\ \Rightarrow\left|x\right|=1\)
Vậy \(\left|x\right|=1\)