\(\left|2x-3\right|-x=\left|2-x\right|\left(\circledast\right)\)
TH1: \(x< \dfrac{3}{2}\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=3-2x\\\left|2-x\right|=2-x\end{matrix}\right.\)
Pt (*) trở thành:
\(3-2x-x=2-x\\ \Leftrightarrow-2x-x+x=2-3\\ \Leftrightarrow-2x=-1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\)
TH2: \(x>2\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|2-x\right|=x-2\end{matrix}\right.\)
Pt (*) trở thành:
\(2x-3-x=x-2\\ \Leftrightarrow2x-x-x=-2+3\\ \Leftrightarrow0x=1\left(ktm\right)\)
TH3: \(\dfrac{3}{2}\le x\le2\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|2-x\right|=2-x\end{matrix}\right.\)
Pt (*) trở thành:
\(2x-3-x=2-x\\ \Leftrightarrow2x-x+x=2+3\\ \Leftrightarrow2x=5\\ \Leftrightarrow x=\dfrac{5}{2}\left(ktmđk\right)\)
Vậy \(x=\dfrac{1}{2}\) duy nhất thỏa mãn phương trình.
