Điều kiện xác định:
$\left\{ \begin{array}{l} {x^2} - 4x + 4 \ge 0\\ 4{x^2} - 12x + 9 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\left( {x - 2} \right)^2} \ge 0\\ {\left( {2x - 3} \right)^2} \ge 0 \end{array} \right. \Rightarrow x \in \mathbb{R}$
$\begin{array}{l} \sqrt {{x^2} - 4x + 4} = \sqrt {4{x^2} - 12x + 9} \\ \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt {{{\left( {2x - 3} \right)}^2}} \\ \Leftrightarrow \left| {x - 2} \right| = \left| {2x - 3} \right|\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 2x - 3\\ x - 2 = 3 - 2x \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = \dfrac{5}{3} \end{array} \right.\\ \Rightarrow S = \left\{ {1;\dfrac{5}{3}} \right\} \end{array}$
