\(a,\left|-4x\right|=x+2\\ TH_1:x\le0\\ -4x=x+2\Leftrightarrow-4x-x=2\Leftrightarrow-5x=2\Leftrightarrow x=-\dfrac{2}{5}\left(tm\right)\\ TH_2:x>0\\ 4x=x+2\Leftrightarrow3x=2\Leftrightarrow x=\dfrac{2}{3}\left(tm\right)\)
Vậy \(S=\left\{-\dfrac{2}{5};\dfrac{2}{3}\right\}\)
\(b,\left|2-x\right|=2-3x\\ TH_1:x\le2\\ 2-x=2-3x\Leftrightarrow2x=0\Leftrightarrow x=0\left(tm\right)\\ TH_2:x>2\\ -2+x=2-3x\Leftrightarrow4x=4\Leftrightarrow x=1\left(ktm\right)\)
Vậy \(S=\left\{0\right\}\)
\(c,\left|2x-3\right|=5x-6\\ TH_1:x\ge\dfrac{3}{2}\\ 2x-3=5x-6\Leftrightarrow-3x=-3\Leftrightarrow x=1\left(ktm\right)\\ TH_2:x< \dfrac{3}{2}\\ -2x+3=5x-6\Leftrightarrow-7x=-9\Leftrightarrow x=\dfrac{9}{7}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{9}{7}\right\}\)
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