$HPT \Leftrightarrow \left\{ \begin{array}{l} x+\sqrt{x^2-2x+5}=3y+\sqrt{y^2+4} (1)\\ x^2-y^2-3x+3y+1=0 (2) \end{array} \right.$Lấy $(1)+(2) \Leftrightarrow x^2-2x+1+\sqrt{x^2-2x+5}=y^2+\sqrt{y^2+4}$
$\Leftrightarrow (x-1)^2+\sqrt{(x-1)^2+4}=y^2+\sqrt{y^2+4}$
$\Leftrightarrow (x-1)^2+4+\sqrt{(x-1)^2+4}=y^2+4+\sqrt{y^2+4}$
$\Leftrightarrow [(x-1)^2+4-y^2-4]+(\sqrt{(x-1)^2+4}-\sqrt{y^2+4})=0$
Áp dụng HĐT, có
$(\sqrt{(x-1)^2+4}-\sqrt{y^2+4})(\sqrt{(x-1)^2+4}+\sqrt{y^2+4}+1)=0$
$\Leftrightarrow \sqrt{x-1)^2+4}=\sqrt{y^2+4}$
$\Leftrightarrow (x-1)^2=y^2$
$* x-1=y$
Thế vào $(2) \Rightarrow \left\{ \begin{array}{l} x=\frac{3}2\\ y=\frac{1}2 \end{array} \right.$
$*x-1=-y \Leftrightarrow x=1-y$
Thế vào $(2) \Rightarrow \left\{ \begin{array}{l} x=\frac{3}4\\ y=\frac{1}4 \end{array} \right.$