2) $1 + 2\sqrt{x^2 - 9x + 18} = x + \sqrt{x^2 - 14x + 33}$
ĐKXĐ: $x \le 3$ hoặc $x \ge 11$
PT $\leftrightarrow 2(\sqrt{x^2 - 9x + 18} - x) = \sqrt{x^2 - 14x + 33} - (x + 1)$
$\leftrightarrow 2.\dfrac{-9(x - 2)}{\sqrt{x^2 - 9x + 18} + x} = \dfrac{-16(x - 2)}{\sqrt{x^2 - 14x + 33} + (x + 1)}$
*TH1: x = 2
*TH2: $-18\sqrt{x^2 - 14x + 33} + 16\sqrt{x^2 - 9x + 18} = 2x + 18$
Kết hợp với pt đã cho ta có hệ tạm:
$\left\{\begin{matrix} -18\sqrt{x^2-14x+33} + 16\sqrt{x^2 - 9x + 18} = 2x + 18 & & \\ \ \sqrt{x^2 - 14x + 33} - 2\sqrt{x^2 - 9x + 18} = 1 - x \end{matrix}\right.$
$\leftrightarrow \sqrt{x^2 - 14x + 33} = \dfrac{3x - 13}{5}$
$\leftrightarrow x = \dfrac{17 + 5\sqrt{5}}{2}$