\(\left|x-3,2\right|+\left|\dfrac{2x-1}{5}\right|=x+3\) (1)
TH1: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow x-3,2+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow5x-16+2x-1=5x+15\Rightarrow2x=32\Leftrightarrow x=16\left(tm\right)\)
TH2: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\((1)\)\(\Rightarrow x-3,2+\dfrac{1-2x}{5}=x+3\Rightarrow5x-16+1-2x=5x+15\)
\(\Rightarrow-2x=0\Rightarrow x=0\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3,2-x+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow16-5x+2x-1=5x+15\Rightarrow8x=0\Leftrightarrow x=0\left(l\right)\)
TH4: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3,2-x+\dfrac{1-2x}{5}=x+3\)
\(\Rightarrow16-5x+1-2x=5x+15\Rightarrow12x=2\Rightarrow c=\dfrac{1}{6}\left(tm\right)\)
Vậy \(x=\left\{16;\dfrac{1}{6}\right\}\)
