ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=m\\a^2+b^2-ab=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=m\\\left(a+b\right)^2-3ab=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=m\\ab=\frac{m^2-m}{3}\end{matrix}\right.\)
Hệ đã cho có nghiệm khi và chỉ khi: \(t^2-m.t+\frac{m^2-m}{3}=0\) có 2 nghiệm ko âm
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=m^2-\frac{4}{3}\left(m^2-m\right)\ge0\\t_1+t_2=m\ge0\\t_1t_2=\frac{m^2-m}{3}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m-m^2\ge0\\m\ge0\\m\left(m-1\right)\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}0\le m\le4\\m\ge0\\\left[{}\begin{matrix}m\le0\\m\ge1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=0\\1\le m\le4\end{matrix}\right.\)