\(1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7x}{3}}{2}\\ \Leftrightarrow1-\dfrac{\dfrac{3x-1-x}{3}}{3}=\dfrac{x}{2}-\dfrac{\dfrac{6x-10+7x}{3}}{2}\\ \Leftrightarrow1-\dfrac{2x-1}{9}=\dfrac{x}{2}-\dfrac{13x-10}{6}\\ \Leftrightarrow\dfrac{9-2x+1}{9}=\dfrac{3x-13x+10}{6}\\ \Leftrightarrow\dfrac{10-2x}{9}=\dfrac{-10x+10}{6}\\ \Leftrightarrow6\left(10-2x\right)=9\left(-10x+10\right)\\ \Leftrightarrow60-12x=-90x+90\\ \Leftrightarrow-90x+90+12x-60=0\\ \Leftrightarrow-78x+30=0\\ \Leftrightarrow x=\dfrac{5}{13}\)
Vậy `S={5/13}`
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