a) \(\left|0,2x-3,1\right|=6,3\)
th1: \(0,2x-3,1\ge0\Leftrightarrow0,2x\ge3,1\Leftrightarrow x\ge\dfrac{3,1}{0,2}=15,5\)
\(\Rightarrow\left|0,2x-3,1\right|=6,3\Leftrightarrow0,2x-3,1=6,3\Leftrightarrow0,2x=6,3+3,1\)
\(\Leftrightarrow0,2x=9,4\Leftrightarrow x=\dfrac{9,4}{0,2}=47\left(tmđk\right)\)
th2: \(0,2x-3,1< 0\Leftrightarrow0,2x< 3,1\Leftrightarrow x< \dfrac{3,1}{0,2}=15,5\)
\(\Rightarrow\left|0,2x-3,1\right|=6,3\Leftrightarrow-\left(0,2x-3,1\right)=6,3\)
\(\Leftrightarrow-0,2x+3,1=6,3\Leftrightarrow0,2x=3,1-6,3\Leftrightarrow0,2x=-3,2\)
\(\Leftrightarrow x=\dfrac{-3,2}{0,2}=-16\left(tmđk\right)\)
vậy \(x=47;x=-16\)
