Đk $x \le 10,y \le 9,2x+6 \ge y,y+11 \ge 2x$$pt(1)\Leftrightarrow (53-5x)\sqrt{10-x}=(48-5y)\sqrt{9-y}$
$\Leftrightarrow \Big[3+5(10-x) \Big]\sqrt{10-x}=\Big[3+5(9-y) \Big]\sqrt{9-y}$
$\Leftrightarrow5\Big( \sqrt{10-x}^3-\sqrt{9-y}^3 \Big)+(\sqrt{10-x}-\sqrt{9-y})=0$
$\Leftrightarrow \Big(\sqrt{10-x}-\sqrt{9-y} \Big)\Big[...\Big]=0\Leftrightarrow y=x-1$
Thế vào $pt(2):\sqrt{x+7}+x^2=\sqrt{10-x}+2x+66$
$\Leftrightarrow \Big(\sqrt{x+7}-4 \Big)+\Big((x-1)^2-64\Big)=\sqrt{10-x}-1\Leftrightarrow x=9$
$\Rightarrow (x,y)=(9,8)$