\(\begin{array}{l}
1)\quad |2x -3| - x = |2-x|\qquad (1)\\
+)\quad x >2\\
(1)\Leftrightarrow 2x - 3 - x = x - 2\\
\Leftrightarrow x - 3 = x - 2\\
\Leftrightarrow -3 = -2\quad \text{(vô lí)}\\
+)\quad \dfrac{3}{2}\leqslant x \leqslant 2\\
(1)\Leftrightarrow 2x - 3 - x = 2 - x\\
\Leftrightarrow 2x = 5\\
\Leftrightarrow x = \dfrac52\quad (loại)\\
+)\quad x < \dfrac32\\
(1)\Leftrightarrow 3 - 2x - x = 2 - x\\
\Leftrightarrow 1 = 2x\\
\Leftrightarrow x = \dfrac12\quad (nhận)\\
\text{Vậy}\ S = \left\{\dfrac12\right\}\\
2)\quad 2|x-3| - |4x-1| = 0\\
\Leftrightarrow 2|x-3| = |4x-1|\\
\Leftrightarrow\left[\begin{array}{l}2(x-3) = 4x - 1\\2(x-3) = - 4x +1\end{array}\right.\\
\Leftrightarrow\left[\begin{array}{l}2x = -5\\6x=7\end{array}\right.\\
\Leftrightarrow\left[\begin{array}{l}x = -\dfrac52\\x=\dfrac76\end{array}\right.\\
\text{Vậy}\ S = \left\{-\dfrac52;\dfrac76\right\}\\
3)\quad |x+2| + |2y + 3| = 0\qquad (3)\\
\text{Ta có:}\\
\begin{cases}|x+2| \geqslant 0\quad\forall x\\|2y + 3| \geqslant 0\quad \forall y\end{cases}\\
\text{Do đó:}\\
(3)\Leftrightarrow |x+2| = |2y + 3| = 0\\
\Leftrightarrow \begin{cases}x +2= 0\\2y + 3 =0\end{cases}\\
\Leftrightarrow \begin{cases}x = -2\\y = -\dfrac32\end{cases}\\
\text{Vậy}\ (x;y) = \left(-2;-\dfrac32\right)
\end{array}\)
