Đk $ x,y \geq 0$ hpt $\Leftrightarrow \begin{cases}x+y+\sqrt{2xy}=4 \\x+y+6+2\sqrt{(x+3)(y+3)}=10 \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ \sqrt{(x+3)(y+3)} - \sqrt{xy}=3 \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ xy+3(x+y)+9=9+xy+6\sqrt{xy} \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ (\sqrt{x}-\sqrt{y})^2=0 \end{cases}\Leftrightarrow \begin{cases}x=y \\ x+x+2\sqrt{x.x}=4 \end{cases}\Leftrightarrow \color{red}{x=y=1}$
Đk $ x,y \geq 0$ hpt $\Leftrightarrow \begin{cases}x+y+
2\sqrt{xy}=4 \\x+y+2\sqrt{(x+3)(y+3)}=10 \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ \sqrt{(x+3)(y+3)} - \sqrt{xy}=3 \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ xy+3(x+y)+9=9+xy+6\sqrt{xy} \end{cases}$$\Leftrightarrow \begin{cases}x+y+2\sqrt{xy}=4 \\ (\sqrt{x}-\sqrt{y})^2=0 \end{cases}\Leftrightarrow \begin{cases}x=y \\ x+x+2\sqrt{x.x}=4 \end{cases}\Leftrightarrow \color{red}{x=y=1}$