Đặt $x^2+2=a(a\ge 0), b=5x-4$
$\Rightarrow x^2+5x-2=x^2+2+5x-4=a+b$
Phương trình trở thành:
$\begin{array}{l} {\left( {a + b} \right)^2} = 4ab\\ \Leftrightarrow {a^2} + {b^2} + 2ab - 4ab = 0\\ \Leftrightarrow {a^2} + {b^2} - 2ab = 0\\ \Leftrightarrow {\left( {a - b} \right)^2} = 0\\ \Leftrightarrow a = b\\ \Leftrightarrow {x^2} + 2 = 5x - 4\\ \Leftrightarrow {x^2} - 5x + 6 = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 3 \end{array} \right.\\ \Rightarrow S = \left\{ {2;3} \right\} \end{array}$
