DKXD : $y \le 1 ,3x+22 \ge xy$
$(1)\Leftrightarrow \sqrt{x^2+9y^2+6xy-4x-12y+8}-\sqrt{(x-y)^2+4}+2(x+y-1)=0$
$\Leftrightarrow \frac{4(2y-1)(x+y-1)}{\sqrt{(x+3y-2)^2+4}+\sqrt{(y-x)^2+4}}+2(x+y-1)=0$
$\Leftrightarrow (x+ý-1).f(x)=0 $
Với $y \ge \frac 12$ thì $f(x)>0$
Với $y < \frac 12$ thì $f(x) \overset{min-cop-xki}\ge \frac{4(2y-1)}{\sqrt{(4y-2)^2+16}}+2=\frac{2y-1}{\sqrt{(2y-1)^2+4}}+2>0$
Vậy $(1)\Leftrightarrow y=1-x$
Thế xuống pt thứ 2
$\sqrt{3x-x(1-x)+22}-\sqrt x=x^2-2+2x+3$
$\Leftrightarrow \sqrt{x^2+2x+22}-\sqrt x=(x+1)^2$
$\Leftrightarrow \sqrt{x^2+2x+22}-5+1-\sqrt x=(x+1)^2-4$
$\Leftrightarrow \frac{(x-1)(x+3)}{\sqrt{x^2+2x+22}+5}=(x-1)(x+3)+\frac{x-1}{\sqrt x+1 }$
$\Leftrightarrow x=1\Leftrightarrow \begin{cases}x=1 \\ y=0 \end{cases}$