\(\Delta=\left(m+1\right)^2-2m=m^2+1>0\Rightarrow\) pt luôn có 2 nghiệm phân biệt
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)
Từ điều kiện bài toán \(\Rightarrow\left\{{}\begin{matrix}x_1\ge0\\x_2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1+x_2\ge0\\x_1x_2\ge0\end{matrix}\right.\) \(\Rightarrow m\ge0\)
\(\sqrt{x_1}+\sqrt{x_2}\le2\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}\le4\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{2m}\le4\Leftrightarrow1-m\ge\sqrt{2m}\)
\(\Rightarrow\left\{{}\begin{matrix}0\le m\le1\\m^2-2m+1\ge2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}0\le m\le1\\m^2-4m+1\ge0\end{matrix}\right.\)
\(\Rightarrow0\le m\le2-\sqrt{3}\)
