\(đặt:\sqrt{x^2+y}=a;\sqrt{x^2+3}=b\left(a\ge0;b>0\right)\)\(\left(x^2+y\ge0\right)\)
\(\left\{{}\begin{matrix}\left(\sqrt{x^2+y}+\sqrt{x^2+3}\right)x=y-3\left(1\right)\\\sqrt{x^2+y}+\sqrt{x}=x+3\left(2\right)\end{matrix}\right.\)
\(\Rightarrow a^2-b^2=x^2+y-x^2-3=y-3\)
\(\Rightarrow pt\left(1\right)\Leftrightarrow\left(a+b\right)x=a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\Leftrightarrow x\left(a+b\right)-\left(a+b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(x-a+b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\Rightarrow\sqrt{x^2+y}=-\sqrt{x^2+3}< 0\left(loại\right)\\x=a-b\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\sqrt{x^2+y}-3+\sqrt{x}-x=0\)
\(\Leftrightarrow a+\sqrt{a-b}=a-b+3\Leftrightarrow\sqrt{a-b}=3-b\Leftrightarrow\left\{{}\begin{matrix}b\le3\\a-b=\left(3-b\right)^2\end{matrix}\right.\)
\(\Rightarrow a-b=\left(3-b\right)^2\Rightarrow x=\left(3-b^2\right)\Rightarrow\sqrt{x^2+3}=b\Leftrightarrow\sqrt{\left(3-b\right)^4+3}=b\Leftrightarrow b=2\left(thỏa\right)\Rightarrow x=1\Rightarrow y=8\left(thỏa\right)\)
\(\)
